2834 斐波那契数

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int f[10110]={0,1,1},n,p,m;
int main(){
    scanf("%d%d%d",&n,&p,&m);
    if(n==0||n==1){printf("%d\n",f[n]);return 0;}
    for(int i=2;i<=m;i++){
        f[i]=(f[i-1]%p+f[i-2]%p)%p;//同余与模算术 
        if(f[i]%p==n){printf("%d\n",i);return 0;}
    } 
    printf("-1\n");
    return 0;
}