AC code
#include<iostream>
using namespace std;
bool isLeapYear(int Y)
{
return Y % 4 == 0 && Y % 100 != 0 || Y % 400 == 0;
}
int year(int Y, int N)
{
while (N > 1)
{
Y += 4;
if (isLeapYear(Y))
N--;
}
return Y;
}
int main()
{
int m,Y,N;
cin >> m;
while (m--)
{
cin >> Y >> N;
if (isLeapYear(Y))
Y = year(Y, N);
else
{
for (;; Y++)
if (isLeapYear(Y))break;
Y = year(Y, N);
}
cout << Y << endl;
}
return 0;
}
An Easy Task
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20104 Accepted Submission(s): 12867
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.