CDZSC_2015寒假新人（1）——基础 h

最新推荐文章于 2015-01-28 16:27:00 发布

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最新推荐文章于 2015-01-28 16:27:00 发布

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原文链接：http://www.cnblogs.com/guofeng1022/p/4240110.html

本文介绍了一个程序，用于计算从指定年份开始的第N个闰年。通过输入起始年份和目标闰年的序号，程序可以快速找出对应的闰年。

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

3 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.

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 5 using namespace std;
 6 int main()
 7 {
 8     int N,n,m;
 9     scanf("%d",&N);
10     while(N--)
11     {
12         scanf("%d%d",&n,&m);
13         if((n%100==0&&n%400==0)||(n%100!=0&&n%4==0))
14         {
15             m--;
16         }
17         else
18         {
19             while(n--)
20             {
21                 if((n%100==0&&n%400==0)||(n%100!=0&&n%4==0))
22                     break;
23             }
24         }
25         for(int i=n/100+1;i<=(n+4*m)/100;i++)
26         {
27             if(i%4!=0)
28             {
29                 m++;
30             }
31         }
32         printf("%d\n",n+4*m);
33     }
34 }