Educational Codeforces Round 13 A.Johny Likes Numbers B. The Same Calendar

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A. Johny Likes Numbers
time limit per test0.5 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.

Input
The only line contains two integers n and k (1 ≤ n, k ≤ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
input
5 3
output
6
input
25 13
output
26
input
26 13
output
39

题目大意：
给定两个数 n 和 k，让你求的是 >n 的第一个能够被 k 整除的数

解题思路：
首先让 n 模上 k，然后 n + k- (n%k) ，OK

代码：

#include <iostream>using namespace std;int main()
{int n, k;while(cin>>n>>k){cout<<n+k-n%k<<endl;}return 0;
}

B. The Same Calendar
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
The girl Taylor has a beautiful calendar for the year y. In the calendar all days are given with their days of week: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday.

The calendar is so beautiful that she wants to know what is the next year after y when the calendar will be exactly the same. Help Taylor to find that year.
Note that leap years has 366 days. The year is leap if it is divisible by 400 or it is divisible by 4, but not by 100(https://en.wikipedia.org/wiki/Leap_year).
Input
The only line contains integer y (1000 ≤ y < 100’000) — the year of the calendar.
Output
Print the only integer y’ — the next year after y when the calendar will be the same. Note that you should find the first year after y with the same calendar.
Examples
input
2016
output
2044
input
2000
output
2028
input
50501
output
50507
Note
Today is Monday, the 13th of June, 2016.

题目大意：
就是给你一个年份，让你求的是下一次跟当前给的年份完全一样的年份，这里所说的完全一样指的是这一年的第一天是周几与当前年是周几完全一样，而且必须是闰年和闰年或者平年和平年。

解题思路：
首先给一个数ans 让ans == 0，然后从当前的年份一直往上加，如果是闰年的话就加 2 因为366%7==2,否则加1（355%7 == 1），当ans%7==0 的时候同时是不是闰年也跟当前年份一样的时候就退出，输出 i 。
My Code：

#include <iostream>using namespace std;
bool isleap(int y)
{if( (y%400==0) || (y%4==0 && y%100!=0) )return 1;return 0;
}
int main()
{int y;while(cin>>y){int ans = 0, i;for(i=y+1; ; i++){if(isleap(i)){ans += 2;///366%7 == 2ans %= 7;}else{ans++;///365%7 == 1ans %= 7;}if(ans==0 && (isleap(y)==isleap(i)))break;}cout<<i<<endl;}return 0;
}

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