关于HDOJ 1076题(An Easy Task(x后第n个闰年)) 的理解(C/C++)

题目描述

                       An Easy Task

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28473 Accepted Submission(s): 18363

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.

Author
    Ignatius.L

原题链接

More info:Question

Accepted代码

直接暴力解，不超时。另外，闰年的计算按照题目要求的做，即认为满足(Y%4==0 && Y%100!=0)那么Y这年就可以被称为闰年。即不考虑大年份闰年的判别情况

但是实际上，闰年分为普通闰年和世纪闰年。

普通闰年:公历年份是4的倍数的，一般是闰年。（如2004年就是闰年）；

世纪闰年:公历年份是整百数的，必须是400的倍数才是闰年（如1900年不是世纪闰年，2000年是世纪闰年）;

大年份闰年：对于数值很大的年份，这年如果能整除3200，并且能整除172800则是闰年。如172800年是闰年，86400年不是闰年（因为虽然能整除3200，但不能整除172800）

#include<iostream>
using namespace std;
int isloop(int y) {
    if ((y % 4 == 0 && y % 100 != 0) || (y % 400 == 0)) return 1;
    else return 0;
}
int main() {
    int t;
    cin >> t;
    while (t--) {
        int year, n;
        cin >> year >> n;
        int nyear = year,count=0;
        while (count < n) {
            if (isloop(nyear)) {
                count++;
            }
            nyear++;
        }
        cout << nyear-1 << endl;
    }
    return 0;
}