题目自己找吧,不贴了
思路:构造如下矩阵
a0 | a1 | a2 | a3 | a4 | a5 | a6 | a7 | a8 | a9 |
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 |
f(x-1) |
f(x-2) |
f(x-3) |
f(x-4) |
f(x-5) |
f(x-6) |
f(x-7) |
f(x-8) |
f(x-9) |
f(x-10) |
当以上两个矩阵相乘后会得到一个新矩阵
f(x) |
f(x-1) |
f(x-2) |
f(x-3) |
f(x-4) |
f(x-5) |
f(x-6) |
f(x-7) |
f(x-8) |
f(x-9) |
如此这般,你只需要把第一个表格自乘(x-9)次,再与f(9)~f(0)相乘即可
代码:
#include <fstream> #include <cstring> #include <cstdio> #include <algorithm> #include <iostream> #include <vector> #include <cmath> #define MAX 11 //修改矩阵大小上限 using namespace std; int x,k,M;//n为矩阵大小,k为幂,M为模 long long sum; bool flag; struct node{ int kk[MAX][MAX]; }unit,a; void init() { //初始化矩阵 for(int i=1;i<MAX;i++) for(int j=1;j<MAX;j++) { unit.kk[i][j]=(i==j); a.kk[i][j]=0; } } node mul(node aa,node bb) { node cc; for(int i=1;i<MAX;i++) for(int j=1;j<MAX;j++) { cc.kk[i][j]=0; for(int p=1;p<MAX;p++) cc.kk[i][j]=(cc.kk[i][j] + (aa.kk[i][p]*bb.kk[p][j])%M)%M; } return cc; } node pow(node aa,int exp) { node p=aa; node q=unit; while(exp) { if(exp&1) q=mul(q,p); exp/=2; p=mul(p,p); } return q; } void cal() { node ans=pow(a,x-9); sum=0; for(int i=1,j=9;i<=10;i++,j--) { sum=(sum+ans.kk[1][i]*j)%M; } printf("%lld\n",sum); } int main() { init(); while(~scanf("%d%d",&x,&M)) { for(int i=1;i<=10;i++) scanf("%d",&a.kk[1][i]); for(int i=2;i<=10;i++) a.kk[i][i-1]=1; cal(); } return 0; }