C - An Easy Task

C - An Easy Task

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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 && Y0!=0) or Y@0==0.

题意分析：

这道题本身没有什么难度，就是求输入的年份之后第N个闰年是那年。

解题思路：

这道题唯一的问题就是判断闰年，题目的提示中还给出了判断条件。根据判断条件编码就完了，是闰年就加一，最后到N时就结束循环，输出年份。

编码：

#include <bits/stdc++.h> using namespace std; int main() { int y, n, t; cin >> t; while (t--) { cin >> y >> n; while (n > 0) { if (y % 4 == 0 && y % 100!=0) { n--; } else if (y % 400 == 0) { n--; } y++; } cout << --y << endl; } return 0; }

最后：

这道题本身是一次过，但是在测试的时候自己还遇到了一些小问题，年数循环的时候的y++会导致最后的答案总是+1，所以强制输出了--y。