大致题意： 把\(10\)个点的暴力代码和输入数据都给你，让你求出输出数据。

子任务\(1\)

第一个子任务自然是拿来送分用的。。。

容易发现就是一个快速乘的过程啊。

代码如下：

#include
    
     #define Tp template
     
      #define Ts template
      
       #define Reg register#define RI Reg int#define Con const#define CI Con int&#define I inline#define W while#define ull unsigned long long#define Inc(x,y) ((x+=(y))>=X&&(x-=X))using namespace std;ull a,b,c;I ull Qmul(ull x,ull y,ull X) {ull t=0;W(y) y&1&&Inc(t,x),(x<<=1)%=X,y>>=1;return t;}//快速乘int main(){    freopen("program1.in","r",stdin),freopen("program1.out","w",stdout);    RI T=10;W(T--) scanf("%llu%llu%llu",&a,&b,&c),printf("%llu\n",Qmul(a,b,c));return 0;//读入、输出}
      
     
    

运行结果：

112394409044857551029211890206774929963705929664622924206923118271862747952553433038054401599690143521422731182360573749675429767239498441843796

子任务\(2\)

题目中给出的式子是：

b = a + b;a = 2 * b - a + c;c = 2 * b - a + c;

逐一代入化简得：

\[b=a+b\]

\[a=2(a+b)-a+c=a+2b+c\]

\[c=2(a+b)-(a+2b+c)+c=a\]

较显然的矩乘板子题。

代码如下：

#include
    
     #define Tp template
     
      #define Ts template
      
       #define Reg register#define RI Reg int#define Con const#define CI Con int&#define I inline#define W while#define LL long long#define Inc(x,y) ((x+=(y))>=X&&(x-=X))using namespace std;LL n;int X;class Mat//定义矩阵类{    private:        int n,m;LL a[3][3];    public:        I Mat(CI x=0,CI y=0):n(x),m(y){for(RI i=0,j;i^x;++i) for(j=0;j^y;++j) a[i][j]=0;}        I LL *operator [] (CI x) {return a[x];}        I Mat operator * (Mat o) Con//矩阵乘法        {            Mat res(n,o.m);RI i,j,k;            for(i=0;i^n;++i) for(j=0;j^o.m;++j) for(k=0;k^m;++k) Inc(res[i][j],1LL*a[i][k]*o[k][j]%X);            return res;        }        I Mat operator ^ (LL y) Con//矩阵快速幂        {            Mat x=*this,res(n,m);for(RI i=0;i^n;++i) res[i][i]=1;            W(y) y&1&&(res=res*x,0),x=x*x,y>>=1;return res;        }};int main(){    freopen("program2.in","r",stdin),freopen("program2.out","w",stdout);    Mat Base(3,3),res;Base[0][0]=Base[0][2]=Base[1][0]=Base[1][1]=Base[2][0]=1,Base[0][1]=2;//初始化转移矩阵    RI T=10;W(T--) scanf("%lld%d",&n,&X),res=Mat(3,1),res[0][0]=1,        res=(Base^n)*res,printf("%d\n",(res[0][0]-2*res[1][0]+res[2][0]+2LL*X)%X);//输出答案    return 0;}
      
     
    

运行结果：

0196642503252344521605575868689593160821107500137

子任务\(3\)

题意显然是求\(0\sim n\)的\(0\sim4\)次幂和。

套公式直接算即可，注意此处\(0^0\)貌似等于\(1\)。

考虑到\(unsigned\ long\ long\)的自然溢出是向\(2^{64}\)取模，这个数不是质数，求逆元不太方便，因此可以考虑使用\(Python\)。

代码如下：

Fin=open("program3.in","r");Fout=open("program3.out","w");n=(int)(Fin.read());#读入mod=2**64;#定义模数Fout.write((str)((n+1)%mod)+"\n");#0次幂和Fout.write((str)((n+1)%mod)+"\n");Fout.write((str)((n*(n+1)//2)%mod)+"\n");#1次幂和Fout.write((str)((n*(n+1)//2)%mod)+"\n");Fout.write((str)((n*(n+1)*(2*n+1)//6)%mod)+"\n");#2次幂和Fout.write((str)((n*(n+1)*(2*n+1)//6)%mod)+"\n");Fout.write((str)((n*n*(n+1)*(n+1)//4)%mod)+"\n");#3次幂和Fout.write((str)((n*n*(n+1)*(n+1)//4)%mod)+"\n");Fout.write((str)((n*(n+1)*(2*n+1)*(3*n*n+3*n-1)//30)%mod)+"\n");#4次幂和Fout.write((str)((n*(n+1)*(2*n+1)*(3*n*n+3*n-1)//30)%mod)+"\n");

运行结果：

100000000000000110000000000000012538972135152631808253897213515263180828060986703145697282806098670314569728657034226489832243265703422648983224321006725932432013721610067259324320137216

子任务\(4\)

分析代码可见，子任务\(4\)以及后面的子任务\(5\)都是根据一个\(01\)矩阵而产生的共\(3\)个问题。

问题\(1\)

求出有多少对点都为\(1\)。

则我们直接求出为\(1\)的点数\(tot\)，然后答案就为\(tot(tot-1)\)。

问题\(2\)

求出到每个为\(1\)的点最近的为\(0\)的点的距离和。

从\(4\)个方向（左上、右上、左下、右下）进行\(DP\)，然后取\(min\)即可。

以从左上方向右下方的\(DP\)为例，转移方程如下：

\[f_{i,j}=min(f_{i-1,j},f_{i,j-1})+1\]

代码如下

#include
    
     #define Tp template
     
      #define Ts template
      
       #define Reg register#define RI Reg int#define Con const#define CI Con int&#define I inline#define W while#define N 5000#define LL long long#define Gmin(x,y) (x>(y)&&(x=(y)))using namespace std;int n,m,ty,seed,a[N+5][N+5];I int Rand(){    static const int P=1000000007,Q=83978833,R=8523467;    return seed=(1LL*Q*seed%P*seed+R)%P;}I void Init(){    scanf("%d%d%d",&n,&m,&ty);    for(RI i=0,j;i^n;++i) for(j=0;j^m;++j) a[i][j]=(Rand()%8)>0;}I LL count1()//解决问题1{    RI i,j,tot=0;for(i=0;i^n;++i) for(j=0;j^m;++j) tot+=a[i][j];//统计为1的点的个数    return 1LL*tot*(tot-1);//计算并返回答案}int f[N+5][N+5],g[N+5][N+5];I void Clear() {for(RI i=0,j;i^n;++i) for(j=0;j^m;++j) f[i][j]=a[i][j]?n+m:0;}//清空I void DP1() {for(RI i=0,j;i^n;++i) for(j=0;j^m;++j) i&&Gmin(f[i][j],f[i-1][j]+1),j&&Gmin(f[i][j],f[i][j-1]+1);}//从左上向右下DPI void DP2() {for(RI i=0,j;i^n;++i) for(j=m-1;~j;--j) i&&Gmin(f[i][j],f[i-1][j]+1),(j+1)^m&&Gmin(f[i][j],f[i][j+1]+1);}//从右上向左下DPI void DP3() {for(RI i=n-1,j;~i;--i) for(j=0;j^m;++j) (i+1)^n&&Gmin(f[i][j],f[i+1][j]+1),j&&Gmin(f[i][j],f[i][j-1]+1);}//从左下向右上DPI void DP4() {for(RI i=n-1,j;~i;--i) for(j=m-1;~j;--j) (i+1)^n&&Gmin(f[i][j],f[i+1][j]+1),(j+1)^m&&Gmin(f[i][j],f[i][j+1]+1);}//从右下向左上DPI LL count2()//解决问题2{    RI i,j;LL ans=0;for(i=0;i^n;++i) for(j=0;j^m;++j) g[i][j]=n+m;//初始化    for(Clear(),DP1(),i=0;i^n;++i) for(j=0;j^m;++j) Gmin(g[i][j],f[i][j]);//DP    for(Clear(),DP2(),i=0;i^n;++i) for(j=0;j^m;++j) Gmin(g[i][j],f[i][j]);    for(Clear(),DP3(),i=0;i^n;++i) for(j=0;j^m;++j) Gmin(g[i][j],f[i][j]);    for(Clear(),DP4(),i=0;i^n;++i) for(j=0;j^m;++j) Gmin(g[i][j],f[i][j]);    for(i=0;i^n;++i) for(j=0;j^m;++j) ans+=g[i][j];return ans;//统计并返回答案}int main(){    freopen("program4.in","r",stdin),freopen("program4.out","w",stdout);    RI T=10;scanf("%d",&seed);W(T--) Init(),printf("%lld\n",ty?count2():count1());return 0;}
      
     
    

运行结果：

653007686440954521614752122997258603126474661480452490358302405089924804530602583604050911640508835

子任务\(5\)

问题\(3\)

求全\(1\)矩形的个数。

考虑枚举右下角，则该行从左往右所能构成的全\(1\)矩阵高度应该是递增的，如：

0 0 0 1 0 0 10 0 1 0 0 1 10 1 1 1 1 1 11 0 1 1 1 1 1

这个例子中从左往右全\(1\)矩阵高度应为\(0,0,2,2,2,3,4\)，这可以用单调栈维护。

又可以发现，以该位置为右下角的所有全\(1\)矩阵高度和即为以该位置为右下角的全\(1\)矩阵个数。

因此我们只需要在单调栈内元素进出时维护好全\(1\)矩阵高度和即可。

代码如下：

#include
    
     #define Tp template
     
      #define Ts template
      
       #define Reg register#define RI Reg int#define Con const#define CI Con int&#define I inline#define W while#define N 5000#define LL long long#define Gmin(x,y) (x>(y)&&(x=(y)))using namespace std;int n,m,ty,seed,a[N+5][N+5];I int Rand(){    static const int P=1000000007,Q=83978833,R=8523467;    return seed=(1LL*Q*seed%P*seed+R)%P;}I void Init(){    scanf("%d%d",&n,&m);    for(RI i=0,j;i^n;++i) for(j=0;j^m;++j) a[i][j]=(Rand()%8)>0;}int S[N+5],V[N+5],f[N+5][N+5];I LL count3()//解决问题3{    LL ans=0;for(RI i=(S[0]=-1,0),j,t,T;i^n;++i) for(T=t=j=0;j^m;++j)    {        f[i][j]=a[i][j]?(i?f[i-1][j]:0)+1:0;//求出该位置向上的连续的1的个数        W(T&&f[i][j]<=f[i][S[T]]) t-=V[T--];//维护单调栈，注意弹出元素要清除贡献        S[++T]=j,ans+=(t+=(V[T]=(S[T]-S[T-1])*f[i][j]));//加入元素，计算贡献并统计    }return ans;}int main(){    freopen("program5.in","r",stdin),freopen("program5.out","w",stdout);    RI T=10;scanf("%d",&seed);W(T--) Init(),printf("%lld\n",count3());return 0;}
      
     
    

运行结果：

3679878010949703301977835379444881183972917324090457401682783493647857493666110

子任务\(6\)

题目中要求的是\(n\)次\(t=at^2+b(mod\ c)\)后\(t\)的值。

显然其最后必然会出现循环节。

因此我们考虑用\(Brent\)判环来倍增求出循环节，然后就可以快速计算了。

还是比较模板的。

代码如下：

#include
    
     #define Tp template
     
      #define Ts template
      
       #define Reg register#define RI Reg int#define Con const#define CI Con int&#define I inline#define W while#define ull unsigned long long#define hl_AK_NOI trueusing namespace std;ull n,a,b,X;int main(){    freopen("program6.in","r",stdin),freopen("program6.out","w",stdout);    int T=10;W(T--)    {        ull i,j,t=0,tx=0,ty=2;scanf("%llu%llu%llu%llu",&n,&a,&b,&X),i=j=0;        W(hl_AK_NOI)//Brent判环        {            if(j=(a*j*j+b)%X,++tx,i==j) break;//找到循环节            tx==ty&&(tx=0,ty<<=1,i=j);//倍增        }        i=j=0;W(t^tx) j=(a*j*j+b)%X,++t;//找到循环节的起始位置        t=1;W(i^j) i=(a*i*i+b)%X,j=(a*j*j+b)%X,++t;//计算循环节的长度        t=(n-t)%tx+1;W(t) i=(a*i*i+b)%X,--t;//快速求出答案        printf("%llu\n",i);//输出    }return 0;}
      
     
    

运行结果：

51804876086886904817920662124375143633101788912620413421071515259611527534029879930634769551026935915030784632253370939454891639113574848946074153301099871450072879095235900336338336004

子任务\(7\)

字母版的数独，\(DLX\)板子题。

代码如下：

#include
    
     #define Tp template
     
      #define Ts template
      
       #define Reg register#define RI Reg int#define Con const#define CI Con int&#define I inline#define W whileusing namespace std;int Case;char a[20][20];struct Operate {int x,y,v;}p[(1<<12)+5];class DancingLinksX {    private:        int tot,sz[(1<<10)+5],lnk[(1<<12)+5],res[(1<<8)+5];        struct node        {            int x,y,u,d,l,r;            I node(CI X=0,CI Y=0,CI U=0,CI D=0,CI L=0,CI R=0):x(X),y(Y),u(U),d(D),l(L),r(R){}        }O[(1<<14)+5];        I bool Dance(CI x)        {            #define Delete(x)\            {\                O[O[O[x].l].r=O[x].r].l=O[x].l;\                for(RI i=O[x].d;i^x;i=O[i].d) for(RI j=O[i].r;j^i;j=O[j].r)\                    O[O[O[j].u].d=O[j].d].u=O[j].u,--sz[O[j].y];\            }            #define Regain(x)\            {\                for(RI i=O[x].d;i^x;i=O[i].d) for(RI j=O[i].r;j^i;j=O[j].r)\                    O[O[j].u].d=O[O[j].d].u=j,++sz[O[j].y];\                O[O[x].l].r=O[O[x].r].l=x;\            }            if(!O[0].r)            {                RI i,k;for(i=1;i^x;++i) a[p[res[i]].x][p[res[i]].y]=p[res[i]].v;                for(k=1;k<=Case;++k) {for(i=1;i<=16;++i) cout<<(a[i]+1);cout<
       
        sz[i]&&(t=i);            Delete(t);for(i=O[t].d;i^t;i=O[i].d)            {                for(res[x]=O[i].x,j=O[i].r;j^i;j=O[j].r) Delete(O[j].y);                if(Dance(x+1)) return 1;                for(j=O[i].l;j^i;j=O[j].l) Regain(O[j].y);            }Regain(t);return 0;        }    public:        I void Init(CI x)        {            RI i;for(tot=x,i=0;i<=x;++i) O[i]=node(0,i,i,i,i-1,i+1);            O[O[0].l=x].r=0,memset(lnk,-1,sizeof(lnk)),memset(sz,0,sizeof(sz));        }        I void Insert(CI x,CI y)         {            ++sz[y],O[++tot]=node(x,y,y,O[y].d),O[y].d=O[O[y].d].u=tot,             ~lnk[x]?(O[tot].l=lnk[x],O[tot].r=O[lnk[x]].r,O[lnk[x]].r=O[O[lnk[x]].r].l=tot)             :(lnk[x]=O[tot].l=O[tot].r=tot);         }        I void Solve() {!Dance(1)&&(puts("NO SOLUTION."),0);}}DLX;int main(){    freopen("program7.in","r",stdin),freopen("program7.out","w",stdout);     RI Ttot=4,i,j,k,cnt,t=0;W(Ttot--)    {        #define P(x,y) ((x-1<<4)+y)        #define T(x,y) (((x-1>>2)<<2)+(y+3>>2))        for(++Case,DLX.Init(1<<10),cnt=0,i=1;i<=16;++i) scanf("%s",a[i]+1);        for(i=1;i<=16;++i) for(j=1;j<=16;++j) for(k=1;k<=16;++k)         {            if(a[i][j]^'?'&&(a[i][j]&31)^k) continue;            p[++cnt].x=i,p[cnt].y=j,p[cnt].v=64|k,            DLX.Insert(cnt,P(i,j)),DLX.Insert(cnt,P(i,k)+256),            DLX.Insert(cnt,P(j,k)+512),DLX.Insert(cnt,P(T(i,j),k)+768);        }DLX.Solve();    }return 0;}
       
      
     
    

运行结果：

NAPILFJBMHEOGDCKJDCHMNIPAFGKOBLEKLMGAEHONCDBPJFIFEOBCDGKIJLPAMHNOFHLJMBGEKIDCNAPPCEAHLDFBMJNIKGOBNIKOAPEFGCHJLMDMGDJNIKCLOPAFHEBIOBPFKLJHEMGDANCDKJEIGMAOBNCLFPHGMNCEPOHDLAFKIBJAHLFDBCNPIKJEGOMEIFMBHADKPOLNCJGHPADGJFICNBEMOKLLJKOPCNMGAHIBEDFCBGNKOELJDFMHPIABLNOMGAPIHJKDCEFHGIMCFBOAPEDKNLJFCAEDKLJBGNOPHMIDKJPHIENFMLCGBAOIBEAKCFLHNOGJMDPLHFNJAPDCIKMEGOBPDOJNMHGELBFAKICCMGKOBIEJADPFLNHOFCHPNMBKJAELIGDJNLDGECHOBFIMPKAGAPIFDOKLCMNHJBEMEKBALJIGDPHOFCNEIBGLPNMDOHJCAFKKPDCEHGANFILBOJMNJHLBOKFMECAIDPGAOMFIJDCPKGBNEHLBLNOMGAPIHJKDCEFHGIMCFBOAPEDKNLJFCAEDKLJBGNOPHMIDKJPHIENFMLCGBAOIBEAKCFLHNOGJMDPLHFNJAPDCIKMEGOBPDOJNMHGELBFAKICCMGKOBIEJADPFLNHOFCHPNMBKJAELIGDJNLDGECHOBFIMPKAGAPIFDOKLCMNHJBEMEKBALJIGDPHOFCNEIBGLPNMDOHJCAFKKPDCEHGANFILBOJMNJHLBOKFMECAIDPGAOMFIJDCPKGBNEHLFMPCJGLEANDBOHKIOEHGNCAMFKILDJPBDBNIFHKPJOGEMACLKALJOIDBCHMPGNEFEGOHPJBKIFNMADLCALFDIENOHJKCBMGPNKCPMFGDBLEAHIJOJIBMHACLODPGEFNKLDIEAPHGNBCOJKFMMHAODLJCPGFKNBIEGPKFEBMNDIAJLCOHBCJNKOFIMELHPGADIOGKCDEHLABNFPMJPFEBLMIAGCJDKOHNHJDLGNPFKMOICEBACNMABKOJEPHFILDGFMPCJGLEANDBOHKIOEHGNCAMFKILDJPBDBNIFHKPJOGEMACLKALJOIDBCHMPGNEFEGOHPJBKIFNMADLCALFDIENOHJKCBMGPNKCPMFGDBLEAHIJOJIBMHACLODPGEFNKLDIEAPHGNBCOJKFMMHAODLJCPGFKNBIEGPKFEBMNDIAJLCOHBCJNKOFIMELHPGADIOGKCDEHLABNFPMJPFEBLMIAGCJDKOHNHJDLGNPFKMOICEBACNMABKOJEPHFILDGFMPCJGLEANDBOHKIOEHGNCAMFKILDJPBDBNIFHKPJOGEMACLKALJOIDBCHMPGNEFEGOHPJBKIFNMADLCALFDIENOHJKCBMGPNKCPMFGDBLEAHIJOJIBMHACLODPGEFNKLDIEAPHGNBCOJKFMMHAODLJCPGFKNBIEGPKFEBMNDIAJLCOHBCJNKOFIMELHPGADIOGKCDEHLABNFPMJPFEBLMIAGCJDKOHNHJDLGNPFKMOICEBACNMABKOJEPHFILDGLDFABIOEGCKJNMHPJPONCFDMLHABIGKEHMEGLJPKFNOIACDBCIKBANHGPMEDFOJLFBMIJPLHAKGEDNCOACJKODIFNLPHBEMGPGHEMBKNDFCOJILAONLDEAGCIJBMPHFKDJIPHOELCAFNGKBMEOGHPMCABIJKLDNFMFACNKJBOGDLEPIHBKNLIGFDMEHPCAOJILCJGHMPKBNAOFEDGHBFDLNOEPMCKJAINAPOKEBJHDIFMLGCKEDMFCAIJOLGHBPNLDFABIOEGCKJNMHPJPONCFDMLHABIGKEHMEGLJPKFNOIACDBCIKBANHGPMEDFOJLFBMIJPLHAKGEDNCOACJKODIFNLPHBEMGPGHEMBKNDFCOJILAONLDEAGCIJBMPHFKDJIPHOELCAFNGKBMEOGHPMCABIJKLDNFMFACNKJBOGDLEPIHBKNLIGFDMEHPCAOJILCJGHMPKBNAOFEDGHBFDLNOEPMCKJAINAPOKEBJHDIFMLGCKEDMFCAIJOLGHBPNLDFABIOEGCKJNMHPJPONCFDMLHABIGKEHMEGLJPKFNOIACDBCIKBANHGPMEDFOJLFBMIJPLHAKGEDNCOACJKODIFNLPHBEMGPGHEMBKNDFCOJILAONLDEAGCIJBMPHFKDJIPHOELCAFNGKBMEOGHPMCABIJKLDNFMFACNKJBOGDLEPIHBKNLIGFDMEHPCAOJILCJGHMPKBNAOFEDGHBFDLNOEPMCKJAINAPOKEBJHDIFMLGCKEDMFCAIJOLGHBPNLDFABIOEGCKJNMHPJPONCFDMLHABIGKEHMEGLJPKFNOIACDBCIKBANHGPMEDFOJLFBMIJPLHAKGEDNCOACJKODIFNLPHBEMGPGHEMBKNDFCOJILAONLDEAGCIJBMPHFKDJIPHOELCAFNGKBMEOGHPMCABIJKLDNFMFACNKJBOGDLEPIHBKNLIGFDMEHPCAOJILCJGHMPKBNAOFEDGHBFDLNOEPMCKJAINAPOKEBJHDIFMLGCKEDMFCAIJOLGHBPN

子任务\(8\)

这题看起来像是要推式子（实际上我本来就是要这么做的），但是\(hl666\)神仙告诉我，这其实可以用拉格朗日插值。

具体做法就是先暴力求出一开始的几个值，然后插值求出\(n\)位置的答案即可。

很暴力，很直接，但很没问题。

不过要特殊处理第\(1\)个点，容易看出这个点答案就是\(C_n^7\)，因此一开始的几个值它都为\(0\)，较大的又无法暴力求出，所以不能拉格朗日插值，只能直接计算。

代码如下：

#include
    
     #define Tp template
     
      #define Ts template
      
       #define Reg register#define RI Reg int#define Con const#define CI Con int&#define I inline#define W while#define X 1234567891#define LL long long#define ull unsigned LL#define Qinv(x) Qpow(x,X-2)using namespace std;int F[11][12];ull n;I void MakeList(CI n)//暴力求出一开始的几个值{    RI a,b,c,d,e,f,g,q=0,r=0,s=0,t=0,u=0,v=0,w=0,x=0,y=0,z=0;    for(a=1;a<=n;++a) for(b=1;b<=n;++b) for(c=1;c<=n;++c)        for(d=1;d<=n;++d) for(e=1;e<=n;++e) for(f=1;f<=n;++f) for(g=1;g<=n;++g)            a
       
        
         
          
           
            
             
              
               
                
                 
                  >=1;return t;}//快速幂I void Lagrange(int *F,CI l,CI r,Con ull& n)//拉格朗日插值{ if(n<=r) return (void)(printf("%d\n",F[n]));RI i,j,s1,s2;LL res=0; for(i=l;i<=r;++i)//计算 { for(s1=s2=1,j=l;j<=r;++j) i^j&&(s1=1LL*s1*((n-j)%X)%X,s2=1LL*s2*((i-j+X)%X)%X); (res+=1LL*F[i]*s1%X*Qinv(s2)%X)>=X&&(res-=X); }printf("%d\n",res);//输出答案}int main(){ freopen("program8.in","r",stdin),freopen("program8.out","w",stdout); RI i;for(i=4;i<=11;++i) MakeList(i);scanf("%llu",&n);//预处理 RI ans=1;for(i=1;i<=7;++i) ans=1LL*ans*((n-i+1)%X)%X*Qinv(i)%X;printf("%d\n",ans);//特殊处理第1个点 for(i=2;i<=10;++i) Lagrange(F[i],4,11,n);return 0;//拉格朗日插值求出剩余点}
                 
                
               
              
             
            
           
          
         
        
       
      
     
    

运行结果：

10183333909937049341053807588114415198571206214153007674852068624333749902182027578380253986

子任务\(9\)

直接上网找\(MD5\)解密网站或者人类智慧？

这里直接给出答案：

1984123456chenlijie$_$weholdthesetruthsto beselfevident

子任务\(10\)

一点开\(cpp\)文件顿时吓懵了：这是什么鬼？

仔细一看其实也并不是很难。

我们单独考虑每个单词，则主要是要求出每个字母最终对答案的贡献。

因此我们可以设\(f_{i,j}\)为在长度为\(i\)的单词中，第\(j\)个字母最终被调用的次数。

\(DP\)预处理\(f\)，然后求答案时调用即可。

代码如下：

#include
    
     #define Tp template
     
      #define Ts template
      
       #define Reg register#define RI Reg int#define Con const#define CI Con int&#define I inline#define W while#define hl_AK_NOI true#define ull unsigned long longusing namespace std;const string s[4]=//存下题目中给出的字符串{    "W(); C();","ALGORITHM();","A(); QUICK(); BROWN(); FOX(); JUMPS(); OVER(); THE(); LAZY(); DOG();",    "WHEN(); IN(); THE(); COURSE(); OF(); HUMAN(); EVENTS(); IT(); BECOMES(); NECESSARY(); FOR(); ONE(); PEOPLE(); TO(); DISSOLVE(); THE(); POLITICAL(); BANDS(); WHICH(); HAVE(); CONNECTED(); THEM(); WITH(); ANOTHER(); AND(); TO(); ASSUME(); AMONG(); THE(); POWERS(); OF(); THE(); EARTH(); THE(); SEPARATE(); AND(); EQUAL(); STATION(); TO(); WHICH(); THE(); LAWS(); NATURE(); AND(); NATURES(); GOD(); ENTITLE(); THEM(); A(); DECENT(); RESPECT(); TO(); THE(); OPINIONS(); OF(); MANKIND(); REQUIRES(); THAT(); THEY(); SHOULD(); DECLARE(); THE(); CAUSES(); WHICH(); IMPEL(); THEM(); TO(); THE(); SEPARATION();\    WE(); HOLD(); THESE(); TRUTHS(); TO(); BE(); SELFEVIDENT(); THAT(); ALL(); MEN(); ARE(); CREATED(); EQUAL(); THAT(); THEY(); ARE(); ENDOWED(); BY(); THEIR(); CREATOR(); WITH(); CERTAIN(); UNALIENABLE(); RIGHTS(); THAT(); THEY(); ARE(); AMONG(); THESE(); ARE(); LIFE(); LIBERTY(); AND(); THE(); PURSUIT(); OF(); HAPPINESS(); THAT(); TO(); SECURE(); THESE(); RIGHTS(); GOVERNMENTS(); ARE(); INSTITUTED(); AMONG(); THEM(); DERIVING(); THEIR(); JUST(); POWER(); FROM(); THE(); CONSENT(); OF(); THE(); GOVERNED(); THAT(); WHENEVER(); ANY(); FORM(); OF(); GOVERNMENT(); BECOMES(); DESTRUCTIVE(); OF(); THESE(); ENDS(); IT(); IS(); THE(); RIGHT(); OF(); THE(); PEOPLE(); TO(); ALTER(); OR(); TO(); ABOLISH(); IT(); AND(); TO(); INSTITUTE(); NEW(); GOVERNMENT(); LAYING(); ITS(); FOUNDATION(); ON(); SUCH(); PRINCIPLES(); AND(); ORGANIZING(); ITS(); POWERS(); IN(); SUCH(); FORM(); AS(); TO(); THEM(); SHALL(); SEEM(); MOST(); LIKELY(); TO(); EFFECT(); THEIR(); SAFETY(); AND(); HAPPINESS(); PRUDENCE(); INDEED(); WILL(); DICTATE(); THAT(); GOVERNMENTS(); LONG(); ESTABLISHED(); SHOULD(); NOT(); BE(); CHANGED(); FOR(); LIGHT(); AND(); TRANSIENT(); CAUSES(); AND(); ACCORDINGLY(); ALL(); EXPERIENCE(); HATH(); SHOWN(); THAT(); MANKIND(); ARE(); MORE(); DISPOSED(); TO(); SUFFER(); WHILE(); EVILS(); ARE(); SUFFERABLE(); THAN(); T(); RIGHT(); THEMSELVES(); BY(); ABOLISHING(); THE(); FORMS(); TO(); WHICH(); THEY(); ARE(); ACCUSTOMED(); BUT(); WHEN(); A(); LONG(); TRAIN(); OF(); ABUSES(); AND(); USURPATIONS(); PURSUING(); INVARIABLY(); THE(); SAME(); OBJECT(); EVINCES(); A(); DESIGN(); TO(); REDUCE(); THEM(); UNDER(); ABSOLUTE(); DESPOTISM(); IT(); IS(); THEIR(); RIGHT(); IT(); IS(); THEIR(); DUTY(); TO(); THROW(); OFF(); SUCH(); GOVERNMENT(); AND(); TO(); PROVIDE(); NEW(); GUARDS(); FOR(); THEIR(); FUTURE(); SECURITY(); SUCH(); HAS(); BEEN(); THE(); PATIENT(); SUFFERANCE(); OF(); THESE(); COLONIES(); AND(); SUCH(); IS(); NOW(); THE(); NECESSITY(); WHICH(); CONSTRAINS(); THEM(); TO(); ALTER(); THEIR(); FORMER(); SYSTEMS(); OF(); GOVERNMENT(); THE(); HISTORY(); OF(); THE(); PRESENT(); KING(); OF(); GREAT(); BRITAIN(); IS(); USURPATIONS(); ALL(); HAVING(); IN(); DIRECT(); OBJECT(); TYRANNY(); OVER(); THESE(); STATES(); TO(); PROVE(); THIS(); LET(); FACTS(); BE(); SUBMITTED(); TO(); A(); CANDID(); WORLD();\    HE(); HAS(); REFUSED(); HIS(); ASSENT(); TO(); LAWS(); THE(); MOST(); WHOLESOME(); AND(); NECESSARY(); FOR(); THE(); PUBLIC(); GOOD();\    HE(); HAS(); FORBIDDEN(); HIS(); GOVERNORS(); TO(); PASS(); LAWS(); OF(); IMMEDIATE(); AND(); PRESSING(); IMPORTANCE(); UNLESS(); SUSPENDED(); IN(); THEIR(); OPERATION(); TILL(); HIS(); ASSENT(); SHOULD(); BE(); OBTAINED(); AND(); WHEN(); SO(); SUSPENDED(); HE(); HAS(); UTTERLY(); NEGLECTED(); TO(); ATTEND(); THEM();\    HE(); HAS(); REFUSED(); TO(); PASS(); OTHER(); LAWS(); FOR(); THE(); ACCOMMODATION(); OF(); LARGE(); DISTRICTS(); OF(); PEOPLE(); UNLESS(); THOSE(); PEOPLE(); WOULD(); RELINQUISH(); THE(); RIGHT(); OF(); REPRESENTATION(); IN(); THE(); LEGISLATURE(); A(); RIGHT(); INESTIMABLE(); TO(); THEM(); AND(); FORMIDABLE(); TO(); TYRANTS(); ONLY();\    HE(); HAS(); CALLED(); TOGETHER(); LEGISLATIVE(); BODIES(); AT(); PLACES(); UNUSUAL(); UNCOMFORTABLE(); AND(); DISTANT(); FROM(); THE(); DEPOSITORY(); OF(); THEIR(); PUBLIC(); RECORDS(); FOR(); THE(); SOLE(); PURPOSE(); OF(); FATIGUING(); THEM(); INTO(); COMPLIANCE(); WITH(); HIS(); MEASURES();\    HE(); HAS(); DISSOLVED(); REPRESENTATIVE(); HOUSES(); REPEATEDLY(); FOR(); OPPOSING(); WITH(); MANLY(); FIRMNESS(); HIS(); INVASION(); ON(); THE(); RIGHTS(); OF(); THE(); PEOPLE();\    HE(); HAS(); REFUSED(); FOR(); A(); LONG(); TIME(); AFTER(); SUCH(); DISSOLUTION(); TO(); CAUSE(); OTHERS(); TO(); BE(); ELECTED(); WHEREBY(); THE(); LEGISLATIVE(); POWERS(); INCAPABLE(); OF(); ANNIHILATION(); HAVE(); RETURNED(); TO(); THE(); PEOPLE(); AT(); LARGE(); FOR(); THEIR(); EXERCISE(); THE(); STATE(); REMAINING(); IN(); THE(); MEANTIME(); EXPOSED(); TO(); ALL(); THE(); DANGERS(); OF(); INVASION(); FROM(); WITHOUT(); AND(); CONVULSION(); WITHIN();\    HE(); HAS(); ENDEAVORED(); TO(); PREVENT(); THE(); POPULATION(); OF(); THESE(); STATES(); FOR(); THAT(); PURPOSE(); OBSTRUCTING(); THE(); LAWS(); OF(); NATURALIZING(); OF(); FOREIGNERS(); REFUSING(); TO(); PASS(); OTHERS(); TO(); ENCOURAGE(); THEIR(); MIGRATION(); HITHER(); AND(); RAISING(); THE(); CONDITION(); OF(); NEW(); APPROPRIATIONS(); OF(); LANDS();\    HE(); HAS(); OBSTRUCTED(); THE(); ADMINISTRATION(); OF(); JUSTICE(); BY(); REFUSING(); HIS(); ASSENT(); OF(); LAWS(); FOR(); ESTABLISHING(); JUDICIARY(); POWERS();\    HE(); HAS(); MADE(); JUDGES(); DEPENDENT(); ON(); HIS(); WILL(); ALONE(); FOR(); THE(); TENURE(); OF(); THEIR(); OFFICE(); AND(); THE(); AMOUNT(); AND(); PAYMENT(); OF(); THEIR(); SALARY();\    HE(); HAS(); ERECTED(); A(); MULTITUDE(); OF(); NEW(); OFFICERS(); AND(); SENT(); HITHER(); SWARMS(); OF(); OFFICERS(); TO(); HARASS(); OUR(); PEOPLE(); AND(); EAT(); OUT(); OUR(); SUBSTANCES();\    HE(); HAS(); KEPT(); AMONG(); US(); IN(); TIMES(); OF(); PEACE(); STANDING(); ARMIES(); WITHOUT(); THE(); CONSENT(); OF(); OUR(); LEGISLATURES();\    HE(); HAS(); AFFECTED(); TO(); RENDER(); THE(); MILITARY(); INDEPENDENT(); OF(); AND(); SUPERIOR(); TO(); THE(); CIVIL(); POWER();\    HE(); HAS(); COMBINED(); WITH(); OTHERS(); TO(); SUBJECT(); US(); TO(); A(); JURISDICTION(); FOREIGN(); TO(); OUR(); CONSTITUTION(); AND(); UNACKNOWLEDGED(); BY(); OUR(); LAWS(); GIVING(); HIS(); ASSENT(); TO(); THEIR(); ACTS(); OF(); PRETENDED(); LEGISLATION();\    FOR(); QUARTERING(); LARGE(); BODIES(); OF(); ARMED(); TROOPS(); AMONG(); US();\    FOR(); PROTECTING(); THEM(); BY(); A(); MOCK(); TRIAL(); FROM(); PUNISHMENT(); FOR(); ANY(); MURDER(); WHICH(); THEY(); SHOULD(); COMMIT(); ON(); THE(); INHABITANTS(); OF(); THESE(); STATES();\    FOR(); CUTTING(); OFF(); OUR(); TRADE(); WITH(); ALL(); PARTS(); OF(); THE(); WORLD();\    FOR(); IMPOSING(); TAXES(); ON(); US(); WITHOUT(); OUR(); CONSENT();\    FOR(); DEPRIVING(); US(); IN(); MANY(); CASES(); OF(); THE(); BENEFITS(); OF(); TRIAL(); BY(); JURY();\    FOR(); TRANSPORTING(); US(); BEYOND(); SEAS(); TO(); BE(); TRIED(); FOR(); PRETENDED(); OFFENSES();\    FOR(); ABOLISHING(); THE(); FREE(); SYSTEMS(); OF(); ENGLISH(); LAWS(); IN(); A(); NEIGHBORING(); PROVINCE(); ESTABLISHING(); THEREIN(); AN(); ARBITRARY(); GOVERNMENT(); AND(); ENLARGING(); ITS(); BOUNDARIES(); SO(); AS(); TO(); RENDER(); IT(); AT(); ONCE(); AN(); EXAMPLE(); AND(); FIT(); INSTRUMENT(); FOR(); INTRODUCING(); THE(); SAME(); ABSOLUTE(); RULE(); THESE(); COLONIES();\    FOR(); TAKING(); AWAY(); OUR(); CHARTERS(); ABOLISHING(); OUR(); MOST(); VALUABLE(); LAWS(); AND(); ALTERING(); FUNDAMENTALLY(); THE(); FORMS(); OF(); OUR(); GOVERNMENTS();\    FOR(); SUSPENDING(); OUR(); OWN(); LEGISLATURES(); AND(); DECLARING(); THEMSELVES(); INVESTED(); WITH(); POWER(); TO(); LEGISLATE(); FOR(); US(); IN(); ALL(); CASES(); WHATSOEVER();\    HE(); HAS(); ABDICATED(); GOVERNMENT(); HERE(); BY(); DECLARING(); US(); OUT(); OF(); HIS(); PROTECTION(); AND(); WAGING(); WAR(); AGAINST(); US();\    HE(); HAS(); PLUNDERED(); OUR(); SEAS(); RAVAGED(); OUR(); COASTS(); BURNT(); OUR(); TOWNS(); AND(); DESTROYED(); THE(); LIVES(); OF(); OUR(); PEOPLE();\    HE(); IS(); AT(); THIS(); TIME(); TRANSPORTING(); LARGE(); ARMIES(); OF(); FOREIGN(); MERCENARIES(); TO(); COMPLETE(); THE(); WORKS(); OF(); DEATH(); DESOLATION(); AND(); TYRANNY(); ALREADY(); BEGUN(); WITH(); CIRCUMSTANCES(); OF(); CRUELTY(); AND(); PERFIDY(); SCARCELY(); PARALLEL(); IN(); THE(); MOST(); BARBAROUS(); AGES(); AND(); TOTALLY(); UNWORTHY(); THE(); HEAD(); OF(); A(); CIVILIZED(); NATION();\    HE(); HAS(); CONSTRAINED(); OUR(); FELLOW(); CITIZENS(); TAKEN(); CAPTIVE(); ON(); THE(); HIGH(); SEAS(); TO(); BEAR(); ARMS(); AGAINST(); THEIR(); COUNTRY(); TO(); BECOME(); THE(); EXECUTIONERS(); OF(); THEIR(); FRIENDS(); AND(); BRETHREN(); OR(); TO(); FALL(); THEMSELVES(); BY(); THEIR(); HANDS();\    HE(); HAS(); EXCITED(); DOMESTIC(); INSURRECTION(); AMONGST(); US(); AND(); HAS(); ENDEAVORED(); TO(); BRING(); ON(); THE(); INHABITANTS(); OF(); OUR(); FRONTIERS(); THE(); MERCILESS(); INDIAN(); SAVAGES(); WHOSE(); KNOWN(); RULE(); OF(); WARFARE(); IS(); AN(); UNDISTINGUISHED(); DESTRUCTION(); OF(); ALL(); AGES(); SEXES(); AND(); CONDITIONS();\    IN(); EVERY(); STAGE(); OF(); THESE(); OPPRESSIONS(); WE(); HAVE(); PETITIONED(); FOR(); REDRESS(); IN(); THE(); MOST(); HUMBLE(); TERMS(); OUR(); REPEATED(); PETITION(); HAVE(); BEEN(); ANSWERED(); ONLY(); BY(); REPEATED(); INJURY(); A(); PRINCE(); WHOSE(); CHARACTER(); IS(); THUS(); MARKED(); BY(); EVERY(); ACT(); WHICH(); MAY(); DEFINE(); A(); TYRANT(); IS(); UNFIT(); TO(); BE(); THE(); RULER(); OF(); A(); FREE(); PEOPLE();\    NOR(); HAVE(); WE(); BEEN(); WANTING(); IN(); ATTENTION(); TO(); OUR(); BRITISH(); BRETHREN(); WE(); HAVE(); WARNED(); THEM(); FROM(); TIME(); TO(); TIME(); OF(); ATTEMPTS(); BY(); THEIR(); LEGISLATURE(); TO(); EXTEND(); AN(); UNWARRANTABLE(); JURISDICTION(); OVER(); US(); WE(); HAVE(); REMINDED(); THEM(); OF(); THE(); CIRCUMSTANCES(); OF(); OUR(); EMIGRATION(); AND(); SETTLEMENT(); HERE(); WE(); HAVE(); APPEALED(); TO(); THEIR(); NATIVE(); JUSTICE(); AND(); MAGNANIMITY(); AND(); WE(); HAVE(); CONJURED(); THEM(); BY(); THE(); TIES(); OF(); OUR(); COMMON(); KINDRED(); TO(); DISAVOW(); THESE(); USURPATION(); WHICH(); WOULD(); INEVITABLY(); INTERRUPT(); OUR(); CONNECTIONS(); AND(); CORRESPONDENCE(); THEY(); TOO(); HAVE(); BEEN(); DEAF(); TO(); THE(); VOICE(); OF(); JUSTICE(); AND(); OF(); CONSANGUINITY(); WE(); MUST(); THEREFORE(); ACQUIESCE(); IN(); THE(); NECESSITY(); WHICH(); DENOUNCES(); OUR(); SEPARATION(); AND(); HOLD(); THEM(); AS(); WE(); HOLD(); THE(); REST(); OF(); MANKIND(); ENEMIES(); IN(); WAR(); IN(); PEACE(); FRIENDS();\    WE(); THEREFORE(); THE(); REPRESENTATIVES(); OF(); THE(); UNITED(); STATES(); OF(); AMERICA(); IN(); GENERAL(); CONGRESS(); ASSEMBLED(); APPEALING(); TO(); THE(); SUPREME(); JUDGE(); OF(); THE(); WORLD(); FOR(); THE(); RECTITUDE(); OF(); OUR(); INTENTIONS(); DO(); IN(); THE(); NAME(); AND(); BY(); AUTHORITY(); OF(); THE(); GOOD(); PEOPLE(); OF(); THESE(); COLONIES(); SOLEMNLY(); PUBLISH(); AND(); DECLARE(); THAT(); THESE(); UNITED(); STATES(); COLONIES(); AND(); INDEPENDENT(); STATES(); THAT(); THEY(); ARE(); ABSOLVED(); BY(); FROM(); ALL(); ALLEGIANCE(); TO(); THE(); BRITISH(); CROWN(); AND(); THAT(); ALL(); POLITICAL(); CONNECTION(); BETWEEN(); THEM(); AND(); THE(); STATE(); THEY(); HAVE(); FULL(); POWER(); TO(); LEVY(); WAR(); CONCLUDE(); PEACE(); CONTRACT(); ALLIANCES(); ESTABLISH(); COMMERCE(); AND(); TO(); DO(); ALL(); OTHER(); ACTS(); AND(); THINGS(); WHICH(); INDEPENDENT(); STATES(); MAY(); OF(); RIGHT(); DO(); AND(); FOR(); THE(); SUPPORT(); OF(); THIS(); DECLARATION(); WITH(); A(); FIRM(); RELIANCE(); ON(); THE(); PROTECTION(); OF(); DIVINE(); PROVIDENCE(); WE(); MUTUALLY(); PLEDGE(); TO(); EACH(); OTHER(); OUR(); LIVES(); OUR(); FORTUNES(); AND(); OUR(); SACRED(); HONOR();\    A(); QUICK(); BROWN(); FOX(); JUMPS(); OVER(); THE(); LAZY(); DOG();\    ALGORITHM();"};const int Times[4]={1,2,2,5};int n;ull a[30],f[20][20];int main(){    freopen("program10.in","r",stdin),freopen("program10.out","w",stdout);    RI T,i,j,l,r;ull ans;string t;    for(a[0]=i=1;i<=26;++i) for(a[i]=(27-i)*a[i-1],j=0;j
      
     
    

运行结果：

6754098618987872142128911773319475681521289117733194756815214433265847896447980144332658478964479801536387630300016538415363876303000165384153638763030001653841536387630300016538415363876303000165384

附录：答案

最后，我把每个子任务的答案综合了一下，放在下面的链接中：