UVA 11582 Colossal Fibonacci Numbers!
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原则1:n的余数最多有n种,所以最多n^2就会一个循环,就像400前后同一天的星期几也一样,是个至少要满足的循环
n是1~1000,f函数从第3项开始算时间超出,所以直接打表
#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<cmath>using namespace std; const int maxn=1000+5;int ans[maxn][maxn*6],p[maxn]; //p记录周期 int pow_mod(int a,int b,int n){if(b==0) return 1;int x=pow_mod(a,b/2,n);int ans=(int)((long long)(x*x)%n);if(b%2) ans=(ans*a)%n;return ans; }void init(){for(int n=2;n<=1000;n++){ans[n][0]=0;ans[n][1]=1;for(int i=2;;i++){ans[n][i]=(ans[n][i-1]+ans[n][i-2])%n;if(ans[n][i-1]==0&&ans[n][i]==1){p[n]=i-1;break;}}}} int main(){int t;scanf("%d",&t);init();int n;unsigned long long a,b;while(t--){cin>>a>>b>>n;cout<<(a==0||n==1?0:ans[n][pow_mod(a%1000,b%1000,p[n])])<<endl;}return 0;}
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