HDOJ An Easy Task 1076

An Easy Task

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16761    Accepted Submission(s): 10697

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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水题，还以为会超时呢。。

#include<stdio.h>bool leap(int y){    if((y%4==0&&y%100!=0)||(y%400==0)) return true;    else return false;}int main(){      int t,y,b,i,s;      scanf("%d",&t);      while(t--){          scanf("%d%d",&y,&b);          i=0,s=1;          if(!leap(y))          s=0;          while(1){              y++;              if(leap(y))s++;              if(s==b)break;          }          printf("%d\n",y);      }      return 0;}