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);
}
}
}