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HDU-1076-An Easy Task(Debian下水题測试.....)
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.
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
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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<iostream>
#include<cstdio>
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;
}
HDU-1076-An Easy Task(Debian下水题測试.....)
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