矩阵快速幂求斐波拉契数列的第n项

来源：互联网发布：淘宝店没流量编辑：程序博客网时间：2024/05/22 06:53

代码：

#include <cstdio>#include <cstring>#define MOD 1000000007#define LL long long#define N 10struct node{int mat[N][N];};int n;node xx;node mul(node a, node b){node c;memset(c.mat, 0, sizeof(c.mat));for(int i= 0; i< n; i++)for(int j= 0; j< n; j++)for(int k= 0; k< n; k++)c.mat[i][j]+= a.mat[i][k]* b.mat[k][j];return c;}node fun(node a, int b){node ans= xx;node t= a;while(b){if(b&1)ans=mul(ans, t);t= mul(t, t);b/= 2;}return ans;}int main(){memset(xx.mat, 0, sizeof(xx.mat));for(int i= 0; i< N; i++)xx.mat[i][i]= 1;node a;int b;while(scanf("%d %d",&n,&b)!=EOF){for(int i= 0; i< n; i++)for(int j= 0; j< n; j++)scanf("%d",&a.mat[i][j]);node ans= fun(a, b);for(int i= 0; i< n; i++){for(int j= 0; j< n; j++)printf("%d ",ans.mat[i][j]);printf("\n");}}return 0;}

POJ 3070

代码：

#include <cstdio>#include <cstring>#define MOD 10000#define LL long long#define N 10struct node{int mat[N][N];};int n;node xx;node mul(node a, node b){node c;memset(c.mat, 0, sizeof(c.mat));for(int i= 0; i< n; i++)for(int j= 0; j< n; j++)for(int k= 0; k< n; k++)c.mat[i][j]= (c.mat[i][j]+ a.mat[i][k]* b.mat[k][j])%MOD;return c;}node fun(node a, int b){node ans= xx;node t= a;while(b){if(b&1)ans=mul(ans, t);t= mul(t, t);b/= 2;}return ans;}int main(){memset(xx.mat, 0, sizeof(xx.mat));for(int i= 0; i< N; i++)xx.mat[i][i]= 1;node a;int b;a.mat[0][0]= a.mat[0][1]= a.mat[1][0]= 1;a.mat[1][1]= 0;n= 2; while(scanf("%d",&b)!=EOF && b!=-1){if(b== 0){printf("0\n");continue;}if(b== 1 || b== 2){printf("1\n");continue;}node ans= fun(a, b-2);printf("%d\n",(ans.mat[0][0] + ans.mat[0][1])%MOD);}return 0;}

0 0