hdu 1076

An Easy Task

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10133 Accepted Submission(s): 6291

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

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <cstring>
 5 
 6 using namespace std;
 7 
 8 bool leap(int n)
 9 {
10     if (n%400 == 0 || (n%4==0 && n%100!=0))
11         return true;
12     else return false;
13 }
14 
15 int main(void)
16 {
17     int t;
18 #ifndef ONLINE_JUDGE
19     freopen("1076.in", "r", stdin);
20 #endif
21     scanf("%d", &t);
22     while (t--)
23     {
24         int n, y;
25         scanf("%d%d", &y, &n);
26         if (leap(y))
27         {
28             int cnt = 1, end;
29             for (int i = y+4; ; ++i)
30             {
31                 if (leap(i))
32                     cnt++;
33                 if (cnt == n)
34                 {
35                     end = i;
36                     break;
37                 }
38             }
39             printf("%d\n", end);
40         }
41         else
42         {
43             int cnt = 0, end;
44             for (int i = y; ; ++i)
45             {
46                 if (leap(i))
47                     cnt++;
48                 if (cnt == n)
49                 {
50                     end = i;
51                     break;
52                 }
53             }
54             printf("%d\n", end);
55         }
56     }
57 
58     return 0;
59 }

这题不难……