nefuoj67-思维-自动求循环节

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

在设置cnt的时候，不能把他（以后做除数）初始化为0，如果这样就错，虽然他在日后的计算中肯定要被赋值，肯定会被赋值成其他数，但是还是re。。如果不赋值的话就不会错。。。。

#include <iostream>#include <cstdio>#include <cstring>/*自动求循环节。*/using namespace std;int a[501];int cnt;int main(){   int a1,b;int n;    while(~scanf("%d%d%d",&a1,&b,&n))    {  //int cnt=0;        memset(a,0,sizeof(a));        if(b==0&&a1==0&&n==0) break;         if(n<3){            printf("1\n");              continue;         }        a[1]=1;a[2]=1;   for(int i=3;i<=500;i++)     {a[i]=((a1*a[i-1])%7+(b*a[i-2])%7)%7;      if(a[i]==1&&a[i-1]==1)      {        cnt=i-2;        break;}     }    int x=n%cnt;     if(x==0) cout<<a[cnt]<<endl;   else   printf("%d\n",a[x]);   }    return 0;}