ACM HDU 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41179    Accepted Submission(s): 8827

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

 

这题看起来比较简单,但是也容易错的.但是我发现网上很多这道题的解发都是有问题的,有的可以通过，是数据不多的原因。

比如A=7，B=7，n=50，或者51，明显结果应该是0，但是有的结果出来时1的。

#include<stdio.h>
int main()
{
    //freopen("test.in","r",stdin);
    //freopen("test.out","w",stdout);
    int A,B,i;
    long n;
    int f[201];
    f[1]=f[2]=1;
    while(scanf("%d %d %ld",&A,&B,&n))
    {
        if(A==0&&B==0&&n==0) break;
        int cnt=0;
        for(i=3;i<=200;i++)//打表找到周期 
        {
            f[i]=(A*f[i-1]+B*f[i-2])%7;
            if(f[i]==1&&f[i-1]==1)break;
            if(f[i]==0&&f[i-1]==0){cnt=1;break;}//这里有个小陷阱，如果A=7，B=7则后面都为0了 
        }
        if(cnt){printf("0\n");continue;}
        if(i>n){printf("%d\n",f[n]);continue;}
        i-=2;//i为周期
        n%=i;
        if(n==0)n=i;
        printf("%d\n",f[n]);    
    } 
    return 0;   
    
}