斐波那契数列

F[0]=0;

F[1]=1;

F[n]=F[n-1]+F[n-2], for n>1

给出n (0<=n<=10000) 和 m (0<m<10000);求斐波那契数列第n项mod m的值(因为太大了);

#include<stdio.h>

#include<string.h>

// 矩阵快速幂

void xx(longlong a[][3],longlong b[][3],longlong m){

longlong tmp[3][3];

tmp[1][1]= tmp[1][2]= tmp[2][1]= tmp[2][2]=0;

for(int i =1; i <=2; i++){

for(int j =1; j <=2; j++){

for(int k =1; k <=2; k++){

tmp[i][j]+=(a[i][k]%m*b[k][j]%m)%m;

}

}

}

for(int i =1; i <=2; i++)

for(int j =1; j <=2; j++) a[i][j]= tmp[i][j]%m;

}

int main(){

// ans初值是单位矩阵

longlong T, ans[3][3], r[3][3], n, m;

r[1][2]= r[2][1]= r[2][2]= ans[1][1]= ans[2][2]=1;

r[1][1]= ans[1][2]= ans[2][1]=0;

scanf("%lld%lld",&n,&m);

if(!n){printf("%lld\n",0% m);return0;}

if(n ==1){printf("%lld\n",1% m);return0;}

n--;

while(n){

if(n &1) xx(ans, r, m);

n >>=1;

xx(r, r, m);

}

printf("%lld\n", ans[2][2]% m);

return0;

}