Winter-1-F Number Sequence 解题报告及测试数据

Time Limit:1000MS     Memory Limit:32768KB 

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.

Sample Input

1 1 3

1 2 10

0 0 0

Sample Output

2

5 

题解：

​因为n过大，f(n-1)和f(n-2)只有0,1,2,3,4,5,6七种情况，又A和B不变，f(n)由f(n-1)和f(n-2)决定，所以至多计算49次后，必将发生循环，所以计算50次即可。              

以下是代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <string>
int c[1000];
using namespace std;
int main(){
    int a,b,n;
    int t1,t2,t;
    c[1]=c[2]=1;
    while(scanf("%d%d%d",&a,&b,&n)!=EOF && (a || b || n)){
        t1 =t2=1;
        int tag=0,i;
        for(i=3;i<=100;i++){
            t = t2;
            t2 = (a*t + b*t1)%7;
            t1 = t ;
            c[i]=t2;
            if(t1==1 && t2==1)break;//发生循环，就终止。
        }
        c[0]=c[i-2];//将循环的最后一个值赋值给c[0]，避免取余数后判断
        printf("%d
",c[n%(i-2)]);//i-2即周期
    }
}