杭电1076 An Easy Task

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.

//题意：找出第n个闰年是几几年

#include<stdio.h>
#include<string.h>
#define max 100000
int a[max+100],b[max+100],i;
void fun()
{
    int j=0;
    memset(a,0,sizeof(a));
    for(i=1;i<=max;++i)
    {
        if((i%4==0&&i%100!=0)||i%400==0)//闰年标记
        {
            a[i]=1;
            //b[++j]=i;
        }
    }
}
int main()
{
    int t,y,n;
    fun();
    scanf("%d",&t);
    while(t--)
    {        
        scanf("%d%d",&y,&n);
        int num=0;
        for(i=y;;++i)
        {
            if(a[i])++num;
            if(num==n)break;
        }
        printf("%d\n",i);
        /*
        if(a[y])
        {
            for(i=1;i<=max;++i)
            {
                if(b[i]==y)
                {
                    break;
                }
            }
            printf("%d\n",b[i+n-1]);
        }
        else
        {
            while(!a[y])++y;
            for(i=1;i<=1000;++i)
            {
                if(b[i]==y)
                {
                    break;
                }
            }
            printf("%d\n",b[i+n-1]);
        }
        */
    }
    return 0;
}