矩阵裸题hdu 1757 A Simple Math Problem

/*难点在构造矩阵构造好矩阵就可以在原有的模板求矩阵的N次方在乘上初始值我求矩阵快速幂用的是递归*/#include<iostream>#include<cstdio>#include<cstring>#define max_n 10#define n 10using namespace std;struct M {int s[max_n][max_n];};int mod;M mul(M a,M b){int i,j,k;M t;memset(t.s,0,sizeof(t.s));for(i=0;i<n;i++)for(j=0;j<n;j++){for(k=0;k<n;k++){t.s[i][j]=(t.s[i][j]+a.s[i][k]*b.s[k][j])%mod;}}return t;}M pow(M a,int t){if(t==1)return a;else{M b=pow(a,t/2);if(t&1)return mul(mul(b,b),a);elsereturn mul(b,b);}}int main(){int k;while(scanf("%d%d",&k,&mod)!=EOF){M a,b,c,d;int i;int s1[max_n][max_n]={0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0};memcpy(a.s,s1,sizeof(a.s));for(i=0;i<10;i++)scanf("%d",&a.s[0][i]);b=pow(a,k-9);memset(c.s,0,sizeof(c.s));for(i=0;i<10;i++)c.s[i][0]=9-i;//注意这个地方，我错了好长时间。d=mul(b,c);printf("%d\n",d.s[0][0]);//int sum=0;//for(i=0;i<10;i++)//sum+=(sum+b.s[0][i]*(9-i))%mod;//printf("%d\n",sum);}return 0;}