UVa11582 [快速幂]Colossal Fibonacci Numbers!

题意：f[]为斐波拉契数列，要你求f[a^b]%n;

思路：利用斐波拉契的性质，找余数的循环节，若f[i]==1&&f[i-1]==0,则循环节为i-1；

求a^b用快速幂

注：UVAunsigned long long输入输出用 %llu

#include<stdio.h>#include<vector>#include<algorithm>using namespace std;typedef unsigned long long ull;ull qpow(ull a,ull b,ull cnt){    a=a%cnt;    ull ans=1;    while(b>0)    {        if(b%2==1)            ans=a*ans%cnt;        b/=2;        a=a*a%cnt;    }    return ans;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        ull a,b,n;        scanf("%llu%llu%llu",&a,&b,&n);        ull i;        vector<ull>fib;        fib.push_back(0);        fib.push_back(1);        ull cnt=0;        for(i=2;i<=n*n;i++)        {            fib.push_back((fib[i-1]+fib[i-2])%n);            cnt++;            if(fib[i]==1&&fib[i-1]==0)                break;        }        if(n!=1)        printf("%llu\n",fib[qpow(a,b,cnt)]);        else printf("0\n");    }    return 0;}