hdu 1076 An Easy Task(判断闰年)

最新推荐文章于 2022-01-22 23:09:17 发布

黎辰

最新推荐文章于 2022-01-22 23:09:17 发布

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文章标签： C++

本文链接：https://blog.youkuaiyun.com/qq_24451605/article/details/44587923

文章详细解析了一个关于生日派对日期的计算问题，涉及闰年的判断逻辑，通过输入的起始年份和所需计算的闰年序号，计算并输出指定的下一个闰年。代码实现简洁高效，适用于初学者理解闰年规律。

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An Easy Task

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16139 Accepted Submission(s): 10309

Problem 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

2108
1904
43236

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

Author

Ignatius.L

#include <cstdio>

using namespace std;

int t,a,b;

bool check ( int a )
{
    if ( a%100 ) return a%4 == 0;
    else return a%400 == 0;
}

int main ( )
{
    scanf ( "%d" , &t );
    while ( t-- )
    {
        scanf ( "%d%d" , &a , &b );
        int j = 0 , i = a;
        for ( ; j < b ; i++ )
            if ( check ( i ) )
                j++;
        printf ( "%d\n" , i-1 );
    }
}