An Easy Task
An Easy Task
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 1 Accepted Submission(s) : 1
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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.
题意：
给出起始年份Y，让你求第N个闰年的具体年份。
注意：We call year Y a leap year only if (Y%4==0 && Y%100!=0) or Y%400==0.
并不是每4年，一闰年，若成立即：Y%4==0 即可。但要满足这个(Y%4==0 && Y%100!=0)。
代码：
#include<stdio.h>
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++)/* 闰年不是隔四年一循环*/
if((i%4==0&&i%100!=0)||(i%400==0))
count++;/*是闰年就++，等到到了第nth时候就停止，i-1就是要求的年份*/
printf("%d\n",i-1);
}
}
}
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++)/* 闰年不是隔四年一循环*/
if((i%4==0&&i%100!=0)||(i%400==0))
count++;/*是闰年就++，等到到了第nth时候就停止，i-1就是要求的年份*/
printf("%d\n",i-1);
}
}
}
