HDU-------An Easy Task

时间：2023-03-09 08:38:14
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

相关文章