POJ 3070

来源：互联网发布：奥卡姆剃刀知乎编辑：程序博客网时间：2024/05/22 17:25

思路：矩阵快速幂，没什么可说的。

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef struct Matrix{    int m[2][2];    Matrix(){        memset(m, 0, sizeof(m));    }}Matrix;Matrix mtMul(Matrix A, Matrix B){    Matrix tmp;    for(int i = 0;i < 2;i ++)        for(int j = 0;j < 2;j ++)            for(int k = 0;k < 2;k ++){                int t = (A.m[i][k] * B.m[k][j])%10000;                tmp.m[i][j] = (tmp.m[i][j] + t)%10000;              }    return tmp;}Matrix mtPow(Matrix A, int k){    if(k == 1) return A;    Matrix tmp = mtPow(A, k >> 1);    Matrix res = mtMul(tmp, tmp);    if(k & 1) res = mtMul(res, A);    return res;}int main(){    int n;    while(~scanf("%d", &n) && (n+1)){        if(n == 0) printf("0\n");        else{            Matrix M;            M.m[0][0] = M.m[0][1] = M.m[1][0] = 1;            M.m[1][1] = 0;            Matrix tmp = mtPow(M, n);            printf("%d\n", tmp.m[1][0]);        }    }    return 0;}

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