hdu 1076 An Easy Task
时间：2016-04-26 20:23:09
收藏：0
阅读：207
An Easy Task
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 19908 Accepted Submission(s): 12725
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.
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).
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 43236HintWe call year Y a leap year only if (Y%4==0 && Y%100!=0) or Y%400==0.
Author
Ignatius.L
Recommend
Y 开始以后的第N个闰年
闰年为能被4整除不能被100整除 或者被400整除
#include <stdio.h>
int main()
{
int ncase;
scanf("%d",&ncase);
while(ncase--)
{
int year,n;
scanf("%d %d",&year,&n);
for(int i=year;;i++)
{
if((i%4==0&&i%100)||i%400==0)
n--;
if(n==0)
{
printf("%d\n",i);
break;
}
}
}
return 0;
}原文：http://blog.csdn.net/su20145104009/article/details/51244970
评论（0）