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
Hint
題意：給兩個數n，m 找到一個最接近n的m的倍數
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int n, b;
while (~scanf("%d%d", &n, &b))
{
int i;
for (i = 1;; i++)
{
if (n > b*i)
continue;
else
break;
}
int ans = b*i,temp=b*(i-1);
if (ans - n > n - temp)
cout << temp << endl;
else
cout << ans << endl;
}
}
/**********************************************************************
Problem: 1040
User: leo6033
Language: C++
Result: AC
Time:8 ms
Memory:2024 kb
**********************************************************************/