題目：Harry Potter and Cyber Sequence Generator

題意，有兩個容器C1,C2,初始的時候C1中有一個數的值為V，給你K個操作，每次都重復這K個操作N遍，最后問你C2中的數是

? ? ?多少。N<=10^100。

1：循環操作的次數巨大，敏感的想到這是矩陣連乘的題目。

2：K個操作可以得出一個矩陣，N個K操作就是這個矩陣的N次方

3：最后再乘以初始矩陣即可

#include <iostream> #include <string.h> #include <stdio.h>using namespace std; typedef long long LL;const LL MOD=1000000007; const int N=3;struct Matrix {LL m[N][N]; };Matrix per={1,0,0,0,1,0,0,0,1};Matrix multi(Matrix a,Matrix b) {Matrix c;int i,j,k;for(i=0;i<N;i++){for(j=0;j<N;j++){c.m[i][j]=0;for(k=0;k<N;k++){c.m[i][j]+=(a.m[i][k]%MOD*(b.m[k][j]%MOD))%MOD;c.m[i][j]%=MOD;}}}return c; }Matrix matrix_mod(Matrix a,LL k) {Matrix ans=per,p=a;while(k){if(k&1){ans=multi(ans,p);k--;}k>>=1;p=multi(p,p);}return ans; }LL getNum(char *str) {LL ans=0;int len=strlen(str);for(int i=0;i<len;i++)ans=ans*10+str[i]-'0';return ans; }int main() {Matrix ans,tmp1;LL T,V,Q,cas=1;char str[105];char str1[105];char str2[105];char str3[105];cin>>T;while(T--){cin>>V;ans=per;while(cin>>str1){if(str1[0]=='E') break;cin>>str2>>str3;tmp1=per;if(str1[0]=='S'){if(strcmp(str2,"C1,")==0&&strcmp(str3,"C2")==0){tmp1.m[0][0]=0;tmp1.m[1][0]=1;}else if(strcmp(str2,"C2,")==0&&strcmp(str3,"C1")==0){tmp1.m[0][1]=1;tmp1.m[1][1]=0;}else if(strcmp(str2,"C1,")==0&&strcmp(str3,"C1")!=0){tmp1.m[0][0]=0;tmp1.m[2][0]=getNum(str3);}else if(strcmp(str2,"C2,")==0&&strcmp(str3,"C2")!=0){tmp1.m[1][1]=0;tmp1.m[2][1]=getNum(str3);}}if(str1[0]=='A'){if(strcmp(str2,"C1,")==0&&strcmp(str3,"C1")==0)tmp1.m[0][0]=2;else if(strcmp(str2,"C2,")==0&&strcmp(str3,"C2")==0)tmp1.m[1][1]=2;else if(strcmp(str2,"C1,")==0&&strcmp(str3,"C2")==0)tmp1.m[1][0]=1;else if(strcmp(str2,"C2,")==0&&strcmp(str3,"C1")==0)tmp1.m[0][1]=1;else if(strcmp(str2,"C1,")==0)tmp1.m[2][0]=getNum(str3);elsetmp1.m[2][1]=getNum(str3);}if(str1[0]=='M'){if(strcmp(str2,"C1,")==0)tmp1.m[0][0]=getNum(str3);elsetmp1.m[1][1]=getNum(str3);}ans=multi(ans,tmp1);}cin>>Q;cout<<"Case "<<cas++<<":"<<endl;while(Q--){cin>>str;LL ret=0;int len=strlen(str);for(int i=0;i<len;i++){ret=ret*10+str[i]-'0';ret%=(MOD-1);}Matrix final=matrix_mod(ans,ret);LL sum=(final.m[0][1]%MOD*(V%MOD)%MOD+final.m[2][1]%MOD)%MOD;cout<<sum<<endl;}}return 0; }

總結

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