HDU 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 115124    Accepted Submission(s): 27959

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

 

Author

CHEN, Shunbao

 

Source

ZJCPC2004

 

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这道题真的让我觉得有些头大，题意简单明了，就是按照它的公式推出后来的项目，然后提交发现是超时的，这个时候才发现后面的那个n非常大，但是通项，通项又求不出来，只好但是看的mod 7 觉得应该可以找到规律了吧，嗯，然后就找，那如果两个 1 再次出现不久刚好是个循环吗，但到这里还是把问题想的太简单了，提交了一次RE掉了，后来差到题解，才发现还这是不够的如果出现  7 7 n 这种例子，后面将全是0，然后数组不管开多大都会爆炸，所以不应该用题目已经给你了的值，应该根据输入的情况定出检查的两个值。所以修改后应该是推后两位去取值

#include<cstdio>
#define maxn 60
using namespace std;
int f[maxn] = {0,1,1};
int main(){
    int a,b,c;
    while(~scanf("%d %d %d",&a,&b,&c) && a && b && c){
        int cnt;
        for(int i = 3;i < 54;i++){
            f[i] = (f[i-1]*a + f[i-2]*b)%7;
            if(i > 5){                          //取值定在5之上是为了防止前面的重复区段
                 if(f[i-1] == f[3] && f[i] == f[4]){
                    cnt = i - 4;
                    break;
                 }
            }
        }
        if(c > 2){
            printf("%d
",f[(c-3)%cnt + 3]);    //这段代码是为了防止取值取到零
        }else{
            printf("1
");
        }
    }
    return  0;
}