hdu 1076 An Easy Task

An Easy Task

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17573 Accepted Submission(s): 11213

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

2005 25

1855 12

2004 10000

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

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一道和闰年年份有关的题，注意时间的计算，容易出错，不过这道题还好，数据小，可以直接暴力水过。

题意：输入Y，N，Y表示当前的年份，求第N个闰年的年份。（假如Y是闰年，则Y为第一个闰年）

附上代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 int main()
 5 {
 6     int n,a,b;
 7     scanf("%d",&n);
 8     while(n--)
 9     {
10         scanf("%d%d",&a,&b);
11         while(b!=0)
12         {
13             if((a%4==0&&a%100!=0)||a%400==0)
14                 b--;
15             if(!b)
16                 break;
17             a++;
18         }
19         printf("%d
",a);
20 
21     }
22     return 0;
23 }