POJ 3233 矩阵快速幂

Matrix Power Series

Time Limit: 3000MS		Memory Limit: 131072K

DescriptionGiven a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.InputThe 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 n nonnegativeintegers below 32,768, giving A’s elements in row-major order.OutputOutput 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

题意：给出n,k,m,和n*n矩阵,求矩阵a+a^2+...a^k

题解：

| A  E |		| A^n  E+A+A^2+...A^(n-1) |

| 0  E |   的n次方为	| 0		E  	  |

所以补全A矩阵  即扩大到四倍

求新矩阵A的k+1次方  再将又上角的矩阵减去E即可

#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;typedef long long ll;struct matrix{ll a[125][125];};matrix fir,per;ll n,m,k;matrix multi(matrix a,matrix b){//两矩阵相乘matrix ans;for(ll i=1;i<=2*n;i++){for(ll j=1;j<=2*n;j++){ans.a[i][j]=0;for(ll k=1;k<=2*n;k++){ans.a[i][j]=(a.a[i][k]*b.a[k][j]+ans.a[i][j])%m;}}}return ans;}matrix temp;matrix pows(matrix c,ll k){//矩阵快速幂temp=per;while(k){if(k&1){temp=multi(temp,c);}k=k/2;c=multi(c,c);}return temp;}int main(){scanf("%lld%lld%lld",&n,&k,&m);ll i,j;for(i=1;i<=n;i++){//左上角for(j=1;j<=n;j++){scanf("%lld",&fir.a[i][j]);}}for(i=1;i<=2*n;i++){//单位矩阵for(j=1;j<=2*n;j++){per.a[i][j]=(i==j);}}for(i=1;i<=n;i++){//右上角for(j=n+1;j<=2*n;j++){fir.a[i][j]=(i==j-n);}}for(i=n+1;i<=2*n;i++){//左下角for(j=1;j<=n;j++){fir.a[i][j]=0;}}for(i=n+1;i<=2*n;i++){//右下角for(j=n+1;j<=2*n;j++){fir.a[i][j]=(i==j);}}matrix ans=pows(fir,k+1);for(i=1;i<=n;i++){for(j=n+1;j<=2*n;j++){if(i==j-n){ans.a[i][j]-=1;if(ans.a[i][j]<0)ans.a[i][j]+=m;//防止减后为-1}printf("%lld",ans.a[i][j]);if(j==2*n)printf("\n");else printf(" ");}}return 0;}