标签：pac int clu return roc end ++ ane cout
problem description：
A number sequence is defined as follows:
f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A, B, and n, you are to calculate the value of f(n).
Input：
The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
Output:
For each test case, print the value of f(n) on a single line.
#include<iostream>
using namespace std;
int f(int A,int B,int n ){
if(n==1||n==2){
return 1;
}
else return (A*f(A,B,n-1)+B*f(A,B,n-2))%7;
}
int main(){
int A,B,n;
cin>>A>>B>>n;
while(A||B||n){
cout<<f(A,B,n)<<endl;
cin>>A>>B>>n;
}
return 0;
}
解决思路：由于每49个数循环一次
#include<iostream>
using namespace std;
int main(){
int A,B,n;
int f[50];
while(cin>>A>>B>>n&&(A||B||n)){
n=n%49;
f[1]=1;
f[2]=1;
for(int i=3;i<=49;i++){
f[i]= (A * f[i-1] + B * f[i-2])%7;
}
cout<<f[n]<<endl;
}
return 0;
}
标签：pac int clu return roc end ++ ane cout
原文地址：https://www.cnblogs.com/pesuedream/p/11308319.html
