hdu 1076（An Easy Task）

An Easy Task（原网址：http://acm.hdu.edu.cn/showproblem.php?pid=1076）

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17263    Accepted Submission(s): 11014

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

32005 251855 122004 10000

Sample Output

2108190443236

题意：

 给你一个数字t，随后的n行每行都有两个数字y和n，y为一个年份，求y年（包括）后的第n个闰年。（n组数据）

参考代码：

#include<iostream>using namespace std;int main(){    int x,y,n,i,s;    cin>>n;    for(i=0;i<n;i++)    {        cin>>x>>y;        s=0;//计数器清零        while(s<y)//当找到的闰年的个数等于数据所给的闰年个数时退出循环        {            if(x%4==0 && x%100!=0)//判断是否是闰年①                s++;//是闰年计数+1            else if(x%400==0)//判断是否是闰年②                s++;//是闰年计数+1            x++;//当前年份后移一年        }        cout<<x-1<<endl;    }    return 0;}

运行结果：

Accepted107615MS1956K552 B

题解：

易错点1：清零。计数器一定要清零，否则就会一直加下去！！！

易错点2：闰年判断。判断规则：四年一闰，百年不闰，四百年又闰。所以当它是4的倍数而不是100的倍数①时判断为闰年，当它是400的倍数②时也是闰年。

易错点3：x-1。最后输出时一定要-1，因为在找到目标数后退出循环前x++依然执行了，要把这次加的抵消掉。

参考知识：

无。