题目连接:
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
#includeusing namespace std;const int inf = 1e9;int main(){ long long n,k; cin>>n>>k; cout<<(n/k+1)*k<