Description

Most of the time when rounding a given number, it is customary to round to some multiple of a power of 10. However, there is no reason why we cannot use another multiple to do our rounding to. For example, you could round to the nearest multiple of 7, or the nearest multiple of 3.
Given an int n and an int b, round n to the nearest value which is a multiple of b. If n is exactly halfway between two multiples of b, return the larger value.
 

Input

Each line has two numbers n and b,1<=n<=1000000,2<=b<=500
Output

The answer,a number per line.
Sample Input

5 10
4 10

Sample Output

10
0

題意：輸出離n最近的b的倍數
分析：
1. n ≤ b
① n < b/2 → 0
② n ≥ b/2 → b
2. n ＞ b
d = n % b
① d < b/2 → 0+n-d
② d ≥ b/2 → b+n-d

#include<cstdio>
 

#include<iostream>
using namespace std;
typedef long long LL;

int main() {
#ifdef _DEBUG
    freopen("data.in", "r", stdin);
#endif // _DEBUG

    int n, b;
    while (scanf("%d%d", &n, &b) == 2) {
        int d = n % b;
        int ans = n - d + ((2 * d < b) ? 0 : b);
        //注意不能寫成int ans = n - d + ((d < b/2) ? 0 : b)整型變數
 

        printf("%d\n", ans);
    }
    return 0;
}