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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16797 Accepted Submission(s): 10727
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
We call year Y a leap year only if (Y%4==0 && Y%100!=0) or Y%400==0.
方法一：
#include<stdio.h>
int main()
{
int n,i;
scanf("%d",&n);
while(n--)
{
int a,b,i=0,num,f;
scanf("%d%d",&a,&b);
num=0;
for(i=a;b>0;i++)
{
f=num+a;
num++;
if((i%4==0&&i%100!=0)||i%400==0)
{
b--;
}}
printf("%d\n",f);
}
return 0;
}
方法二：
#include<stdio.h>
int main()
{
int n,i;
scanf("%d",&n);
while(n--)
{
int a,b,i=0,num,f;
scanf("%d%d",&a,&b);
while(b!=0)
{
num=a+i;
i++;
if((num%4==0&&num%100!=0)||num%400==0)
{
b--;
}
}
printf("%d\n",num);
}
return 0;
}
An Easy Task
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原文地址：http://blog.csdn.net/l15738519366/article/details/45968697
