An Easy TaskTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17062Accepted 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.
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
32005 251855 122004 10000
Sample Output
2108190443236
Hint
We 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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We have carefully selected several similar problems for you:10211097106110321170题目的意思很简单：就是让你求出从Y开始的年份第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;}
你的脸是为了呈现上帝赐给人类最贵重的礼物–微笑，
