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

 

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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
       
        #include
        
         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;}