程序師世界是廣大編程愛好者互助、分享、學習的平台,程序師世界有你更精彩!
首頁
編程語言
C語言|JAVA編程
Python編程
網頁編程
ASP編程|PHP編程
JSP編程
數據庫知識
MYSQL數據庫|SqlServer數據庫
Oracle數據庫|DB2數據庫
 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> An Easy Task

An Easy Task

編輯:C++入門知識

An Easy Task

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 1   Accepted Submission(s) : 1

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

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

3
2005 25
1855 12
2004 10000

Sample Output

2108
1904
43236

Hint


題意: 給出起始年份Y,讓你求第N個閏年的具體年份。 注意:         並不是每4年,一閏年,若成立即:即可。但要滿足這個 int main(){
    int T,n,y;
    int i,count;
    while(scanf("%d",&T)!=EOF){
    while(T--){
        count=0;
        scanf("%d%d",&y,&n);
        for(i=y;count<n;i++)/* 閏年不是隔四年一循環*/            
            if((i%4==0&&i%100!=0)||(i%400==0))
                count++;/*是閏年就++,等到到了第nth時候就停止,i-1就是要求的年份*/
        printf("%d\n",i-1);
    }
    }
}

  1. 上一頁:
  2. 下一頁:
Copyright © 程式師世界 All Rights Reserved