ACM HDU 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 64006 Accepted Submission(s): 14732

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

 

Recommend

 

JGShining

 

#include<stdio.h>
#include<math.h>
#include<string.h>
int f[1000];
int main()
{
    int a, b;
    int n, i;
    while(1)
    {
        memset(f, 0, sizeof(f));
        scanf("%d%d%d", &a, &b, &n);
        if(a==0 && b==0 && n == 0) break;
        f[0] = f[1] = 1;
        if(n == 1 || n ==2 )
        {
            printf("1\n");
            continue;
        }
        for(i = 2; i<1000; i++)
        {
            f[i] = (a * f[i-1] + b * f[i-2])% 7;
            if(f[i-1] == 1 && f[i-2] == 1 && i != 2  )  break;    
        }
        printf("%d\n",f[(n-1)%(i-2)]);
    }
    return 0;
}

解题报告：

RE(STACK OVERFLOW) 很明显的少做题就吃定了这个亏，当判为RE是我还是不知所措，为何会出现这种情况，天真的我刚开始敲代码时就直接用递归去做，测试数据放上去丝毫没有问题，RE后查看了Discuss，放上大一点的测试数据，愣是几分钟也没有出结果，无奈之下看见了循环的字眼，我就知道我错哪里了，定义数组，通过输出中间值找到循环的条件，提交上去终于AC了，松了一口气，这题可是费了四五个小时，新手上路之寸步难行！