Number Sequence hdu1005

题意

算第n项，f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

分析

可能的情况0，1，2，3，4，5，6，f(n-1),f(n-2)
所以f(n)最多有49种方案有f(n-1),f(n-2)得来
最大循环节49，n%49即可

code

#include<iostream>
#include<algorithm>
#include<map>
#include<string>
using namespace std;
int main(){int f[50];f[1]=f[2]=1;int a,b,n,i;//freopen("in.txt","r",stdin);while(cin>>a>>b>>n&&a!=0&&b!=0&&n!=0){int maxn=n<50?n:49;for(i=3;i<=maxn;i++){f[i]=(a*f[i-1]+b*f[i-2])%7;}cout<<f[n%49]<<endl;}return 0;
}

转载于:https://www.cnblogs.com/mch5201314/p/9813919.html

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Number Sequence hdu1005

题意

分析

code

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