HDU 1005 Number Sequence (数学规律)

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 104190    Accepted Submission(s): 25232

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

 

 

看了大神的代码才知道水题也是要很严谨去对待的啊。

题目主要是要找循环节。

一开始找规律于是又了下面的代码

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int a[100];
int main()
{
    //freopen("in.txt","r",stdin);
    int n,A,B;
    while(scanf("%d %d %d",&A,&B,&n)&&(n||A||B))
    {
        a[1]=1;
        a[2]=1;
        for(int i=3;i<=49;i++)
            a[i]=(A*a[i-1]+B*a[i-2])%7;
        printf("%d\n",a[n%49]);
    }
    return 0;
}

后来发现如果A=7，B=7，n>2的话，输入都应该会是0，但是实际输出为1。虽然数据有点水，所以还是要严谨对待。

做法还是找循环节，只是说多了很多细节，要好好体会。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
const int MAXN=200+10;
int a[MAXN];
int main()
{
    //freopen("in.txt","r",stdin);
    int A,B,n,i;
    a[1]=a[2]=1;
    while(scanf("%d %d %d",&A,&B,&n)!=EOF)
    {
        if(A==B&&B==n&&n==0)break;
        int flag=0;
        for(i=3;i<=200;i++)
        {
            a[i]=(A*a[i-1]+B*a[i-2])%7;
            if(a[i]==1&&a[i-1]==1) 
                break;
            if(a[i]==0&&a[i-1]==0){
                flag=1;break;
            }   
        }
        if(flag)
        {
            printf("0\n");
            continue;
        }
        if(i>n)
        {
            printf("%d\n",a[n]);
            continue;
        }
        i=i-2;
        n=n%i;
        if(n==0)n=i;
        printf("%d\n",a[n]);
    }
    return 0;
}