这道题虽然不是很难（看着好像不难），代码量也不大，但是我经历了超时，超内存才整出来，所以把这题记录下来

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.

Sample Input

1 1 3 1 2 10 0 0 0

Sample Output

2 5

Time Limit Exceeded

#include<cstdio> int main() { int A,B,n; while(scanf("%d%d%d",&A,&B,&n)!=EOF) { if(A==0&&B==0&&n==0) break; int fn_1=1,fn_2=1; int fn_3; for(int i=2;i<n;i++) { fn_3=(A*fn_2+B*fn_1)%7; fn_1=fn_2; fn_2=fn_3; } printf("%d\n",fn_3); } return 0; }

代码思路是简单，就暴力求解……n的数据很大，就会超时

Memory Limit Exceeded

#include<cstdio> #include<vector> using namespace std; int main() { int A,B,n; while(scanf("%d%d%d",&A,&B,&n)!=EOF) { if(A==0&&B==0&&n==0) break; vector<int> m; int fn_1=1,fn_2=1; int fn_3; m.push_back(fn_1);m.push_back(fn_2); for(int i=2;i<n;i++) { fn_3=(A*fn_2+B*fn_1)%7; fn_1=fn_2; fn_2=fn_3; if(fn_2==1&&fn_1==1) { break; } else m.push_back(fn_2); } if(n==1||n==2) printf("1\n"); else printf("%d\n",m[(n-1)%(m.size()-1)]); } return 0; }

用vector来解，空间不够……这个的思路就是求循环的一个周期，跟着计算就可

AC代码

#include<cstdio> int main() { int A,B,n; while(scanf("%d%d%d",&A,&B,&n)!=EOF) { int m[49]; if(A==0&&B==0&&n==0) break; int fn_1=1,fn_2=1; int fn_3; m[0]=1;m[1]=1; for(int i=2;i<49;i++) { fn_3=(A*fn_2+B*fn_1)%7; fn_1=fn_2; fn_2=fn_3; m[i]=fn_2; } printf("%d\n",m[(n-1)%49]); } return 0; }

f(n-1)与f(n-2)的取值范围为[0,6],七种情况，49(7*7)是一个循环（也可能比49小) 只记录49之前的就可