【poj3233】 Matrix Power Series

http://poj.org/problem?id=3233 (题目链接)

题意

　　给出一个n×n的矩阵A，求模m下A+A2+A3+…+Ak 的值

Solution

　　今日考试就A了这一道题。。

　　当k为偶数时，原式=(Ak2+1)×(A1+A2+...+Ak2)。

　　当k为奇数的时候将Ak乘上当前答案后抠出去，最后统计答案时再加上。所以我们就一路快速幂搞过去，AC

代码

// poj3233
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define inf 2147483640
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;

const int maxn=31;
int A[maxn][maxn],B[maxn][maxn],C[maxn][maxn],T[maxn][maxn],tmp[maxn][maxn],ans[maxn][maxn],D[maxn][maxn];
int n,m;

void pow(int k) {
    for (int i=1;i<=n;i++) {
        for (int j=1;j<=n;j++) B[i][j]=0,T[i][j]=A[i][j];
        B[i][i]=1;
    }
    while (k) {
        if (k&1) {
            for (int i=1;i<=n;i++)
                for (int j=1;j<=n;j++) {
                    tmp[i][j]=0;
                    for (int k=1;k<=n;k++) tmp[i][j]=(tmp[i][j]+B[i][k]*T[k][j])%m;
                }
            for (int i=1;i<=n;i++)
                for (int j=1;j<=n;j++) B[i][j]=tmp[i][j];
        }
        for (int i=1;i<=n;i++)
            for (int j=1;j<=n;j++) {
                C[i][j]=0;
                for (int k=1;k<=n;k++) C[i][j]=(C[i][j]+T[i][k]*T[k][j])%m;
            }
        for (int i=1;i<=n;i++)
            for (int j=1;j<=n;j++) swap(C[i][j],T[i][j]);
        k>>=1;
    }
}
void update() {
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++) {
            tmp[i][j]=0;
            for (int k=1;k<=n;k++) tmp[i][j]=(tmp[i][j]+ans[i][k]*B[k][j])%m;
        }
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++) ans[i][j]=tmp[i][j];
}
int main() {
        int k;
    scanf("%d%d%d",&n,&k,&m);
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++) scanf("%d",&A[i][j]);
    memset(ans,0,sizeof(ans));for (int i=1;i<=n;i++) ans[i][i]=1;
    while (k>1) {
        if (k&1) {
            for (int i=1;i<=n;i++)
                for (int j=1;j<=n;j++) T[i][j]=A[i][j];
            pow(k);
            for (int i=1;i<=n;i++)
                for (int j=1;j<=n;j++) {
                    tmp[i][j]=0;
                    for (int k=1;k<=n;k++) tmp[i][j]=(tmp[i][j]+B[i][k]*ans[k][j])%m;
                }
            for (int i=1;i<=n;i++)
                for (int j=1;j<=n;j++) D[i][j]=(D[i][j]+tmp[i][j])%m;
        }
        pow(k/2);
        for (int i=1;i<=n;i++) B[i][i]=(B[i][i]+1)%m;
        update();
        k>>=1;
    }
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++) B[i][j]=A[i][j];
    update();
    for (int i=1;i<=n;i++) {
        for (int j=1;j<=n;j++) printf("%d ",(ans[i][j]+D[i][j])%m);
        printf("
");
    }
    return 0;
}