HDU-------An Easy Task

 

An Easy Task

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4088 Accepted Submission(s): 2327

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

easy task并不easy 无奈 不过不着急慢慢来 在错误和失败中学到东西才是重要的

上代码

#include <iostream>
using namespace std;
//int leapJudge(int);
int main(){
    int cont,sYear,pYear;
    
    cin>>cont;
    //while(cont--)//这个cont--还是谜一样的存在 ！while(i--)中i是有个初值的，每次循环i会减1，当i的值等于0的时候，循环就会终止！
    
    for(int i =0;i<cont;i++)
    {
        int sYear = 0;
        int pYear = 0;
        cin>>sYear>>pYear;
        int leapcont = 0;  //！！！！！！重大错误就出现在这句 之前把他放在第一层循环外 也是脑子瓦特掉了 
            int j;        
            for(j=sYear;;j++){
                if(leapcont==pYear) break;
                if((j%4==0&&j%100!=0)|j%400==0)//去掉了判断函数 因为确实是没什么卵用
                     leapcont++;
            }
        cout<<j-1<<endl;//此处也是有一个错误 应该输出j-1
    } 
    //system("pause");
    return 0;
}
//int leapJudge(int Y){
//     if( (Y%4==0 && Y%100!=0)||Y%400==0)
//    return 1;
//}

//#include<iostream>
//using namespace std;
//
//int main()
//{
//    int cases;//
//    int k;
//    cin>>cases;
//    while(cases--)
//    {
//        int y,n;
//        cin>>y>>n;
//        int num=0;
//        for(k=y;;k++)
//        {
//            if(num==n) break;
//            //判断是不是闰年
//            if(k%4==0 && k%100!=0 || k%400==0)
//                num++;
//        }
//        cout<<k-1<<endl;
//    }
//    system("pause");
//    return 0;
//}

总结：循环之间的逻辑不清晰