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HDOJ 1076

时间:2015-01-27 23:16:30      阅读:273      评论:0      收藏:0      [点我收藏+]

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An Easy Task

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15396    Accepted Submission(s): 9830


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
3 2005 25 1855 12 2004 10000
 

 

Sample Output
2108 1904 43236

 

 

 

代码如下:

 1 #include <iostream>
 2 using namespace std;
 3 
 4 bool is_leapyear(int n)
 5 {
 6     if((n % 4 == 0) && (n % 100 != 0))
 7         return true;
 8     else if(n % 400 == 0)
 9         return true;
10     else if((n % 3200 == 0) && (n % 172800 == 0))
11         return true;
12     else 
13         return false;
14 }
15 
16 int main()
17 {
18     int T,i,start,num;
19     cin>>T;
20     while(T--)
21     {
22         cin>>start>>num;
23         if(is_leapyear(start) == true)
24             i = 1;
25         else 
26             i = 0;
27         start++;
28         while(i < num)
29         {
30             if(is_leapyear(start) == true)
31                 i++;
32             start++;
33         }
34         cout<<start-1<<endl;
35     }
36     return 0;
37 }

 

HDOJ 1076

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原文地址:http://www.cnblogs.com/baichenjia/p/4254569.html

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