题目连接：
http://www.codeforces.com/contest/678/problem/A
Description
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.
Sample Input
5 3
Sample Output
6
Hint
题意
找到一个大于n的最小值p，使得p%k==0
题解：
O1公式题
答案是(n/k+1)k
显然。
代码
#include<bits/stdc++.h>
using namespace std;
const int inf = 1e9;
int main()
{
long long n,k;
cin>>n>>k;
cout<<(n/k+1)*k<<endl;
return 0;
}