HDU-1076-An Easy Task(Debian下水题測试.....)

An Easy Task

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

Total Submission(s): 17062    Accepted Submission(s): 10902

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

 

Recommend

We have carefully selected several similar problems for you:  1021 1097 

pid=1061" target="_blank">1061 1032 1170






题目的意思非常easy：

就是让你求出从Y開始的年份第N个闰年......甚至不用考虑时间复杂度，强行AC！

主要是Debian下命令行不熟悉，费了不少时间......

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int t,Y,N;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&Y,&N);
        while(1)
        {
            if((Y%4==0&&Y%100!=0)||(Y%400==0))
            {
                N--;
            }
            if(N==0)
                break;
            Y++;
        }
        printf("%d\n",Y);
    }
    return 0;
}