HDU 1076 An easy task

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
View Code
 1 #include<iostream>
 2 #include<stdio.h>
 3 using namespace std;
 4 bool judge(int n) {
 5  return (n%4==0&&n%100!=0||n%400==0);
 6 }
 7 int Y,n;
 8 int main() {
 9  int T;
10  scanf("%d",&T);
11  while(T--) {
12   scanf("%d%d",&Y,&n);
13   while(n) {
14    if(judge(Y)) {
15     n--;
16     if(n!=0) 
17     Y+=4;
18    }
19    else
20    Y+=1;
21   }
22   printf("%d\n",Y);
23  }
24   return 0;
25 }
原文地址:https://www.cnblogs.com/xiaxiaosheng/p/3073777.html