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): 17145    Accepted Submission(s): 10946
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
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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 1 #include<math.h>

 2 #include<stdio.h>

 3 #include<string.h>

 4 #include<iostream>

 5 #include<algorithm>

 6 using namespace std;

 7 

 8 bool judge(int Y)

 9 {

10     if( (Y%4==0 && Y%100!=0) || Y%400==0 )return 1;

11     else return 0;

12 }

13 

14 int n,y;

15 

16 int main()

17 {

18     int t;cin>>t;

19     while(t--)

20     {

21         scanf("%d%d",&y,&n);

22         int i=0;

23         while(1)

24         {

25 

26             if(judge(y))i++;

27             if(i==n){printf("%d
",y);break;}

28             y++;

29         }

30     }

31     return 0;

32 }

33 

34 //freopen("1.txt", "r", stdin);

35     //freopen("2.txt", "w", stdout);

36 //**************************************