hdu 1076 An Easy Task
An Easy Task
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17573 Accepted Submission(s):
11213
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
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
一道和闰年年份有关的题,注意时间的计算,容易出错,不过这道题还好,数据小,可以直接暴力水过。
题意:输入Y,N,Y表示当前的年份,求第N个闰年的年份。(假如Y是闰年,则Y为第一个闰年)
附上代码:
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 int main() 5 { 6 int n,a,b; 7 scanf("%d",&n); 8 while(n--) 9 { 10 scanf("%d%d",&a,&b); 11 while(b!=0) 12 { 13 if((a%4==0&&a%100!=0)||a%400==0) 14 b--; 15 if(!b) 16 break; 17 a++; 18 } 19 printf("%d\n",a); 20 21 } 22 return 0; 23 }