HDU 1076 An Easy Task

最新推荐文章于 2019-03-18 20:48:45 发布

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本文探讨了如何计算从给定年份开始的第N个闰年，涉及闰年的数学判定逻辑，通过实例演示了算法实现过程。

An Easy Task

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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Description

Ignatius was born in a leap year, so he want to know when he could hold his birthday party. Can you tell him?

Given a positive integers Y which indicate the start year, and a positive integer N, your task is to tell the Nth leap year from year Y.

Note: if year Y is a leap year, then the 1st leap year is year Y.

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains two positive integers Y and N(1<=N<=10000).

Output

For each test case, you should output the Nth leap year from year Y.

Sample Input

Sample Output

Hint

 We call year Y a leap year only if (Y%4==0 && Y%100!=0) or Y%400==0.

分析

对每个年份进行判定，跑一次循环即可。

AC代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

bool isleapyear(int n)
{
    if(n%400 == 0) return true;
    if(n%100 != 0 && n % 4 == 0) return true;
    return false;
}

int main()
{
    int t,y,n;
    scanf("%d",&t);
    while(t--)
    {
        int ans = 0;
        scanf("%d%d",&y,&n);
        if(isleapyear(y)) ans++;
        int yu = y%4;
        y += 4 - yu;                   //对第一个对4%==0的年份进行判定
        if(isleapyear(y)) ans++;
        while(ans < n)
        {
            y += 4;
            if(isleapyear(y)) ans++;   //对每个年份判定
        }
        printf("%d\n",y);
    }
    return 0;
}