标签：sam 图片 element lib algo img desc putchar pac
Description
Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.
Input
The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 109) and m (m < 104). Then follow n lines each containing nnonnegative integers below 32,768, giving A’s elements in row-major order.
Output
Output the elements of S modulo m in the same way as A is given.
Sample Input
2 2 4 0 1 1 1
Sample Output
1 2 2 3
Source
求出Sk+1再减去I即可。
代码：
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; int n,k,m; void mul(int (*a)[60],int (*b)[60]) { int d[60][60] = {0}; for(int i = 0;i < n * 2;i ++) { for(int j = 0;j < n * 2;j ++) { for(int k = 0;k < n * 2;k ++) { d[i][j] = (d[i][j] + a[i][k] * b[k][j]) % m; } } } for(int i = 0;i < n * 2;i ++) { for(int j = 0;j < n * 2;j ++) { a[i][j] = d[i][j]; } } } void getans(int (*a)[60]) { int d[60][60]; for(int i = 0;i < n * 2;i ++) { for(int j = 0;j < n * 2;j ++) { d[i][j] = a[i][j]; } } while(k) { if(k % 2)mul(d,a); k /= 2; mul(a,a); } for(int i = n;i < n * 2;i ++) { for(int j = 0;j < n;j ++) { if(j)putchar(‘ ‘); printf("%d",i == j + n ? (d[i][j] + m - 1) % m : d[i][j]); } putchar(‘\n‘); } } int main() { int a[60][60] = {0}; scanf("%d%d%d",&n,&k,&m); for(int i = 0;i < n;i ++) { for(int j = 0;j < n;j ++) { scanf("%d",&a[i][j]); } } for(int i = n;i < n * 2;i ++) { a[i][i - n] = a[i][i] = 1; } getans(a); }
标签：sam 图片 element lib algo img desc putchar pac
原文地址：https://www.cnblogs.com/8023spz/p/9433002.html
