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 =;i<cont;i++)
     {
         int sYear = ;
         int pYear = ;
         cin>>sYear>>pYear;
         int leapcont = ;  //！！！！！！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%==&&j%!=)|j%==)//去掉了判断函数 因为确实是没什么卵用
                      leapcont++;
             }
         cout<<j-<<endl;//此处也是有一个错误 应该输出j-1
     }
     //system("pause");
     return ;
 }
 //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