Round-number(进位题)

JakeRame

于 2018-06-04 20:56:51 发布

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本文介绍了一种基于特定基数b的进位算法实现，对于给定整数n，该算法将其调整为最接近的b的倍数。若n恰好位于两个b倍数中间，则选择较大的那个倍数。

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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

题目含义：n在b的基础上进位，即((n%b)>=b/2)进位，否则就舍去。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath> 
using namespace std;
int main(){
	int n,b;
	while(~scanf("%d %d",&n,&b)){
		if(2*(n%b)>=b){
			cout<<n-n%b+b<<endl;
		}
		else{
			cout<<n-n%b<<endl;
		}
	}
	return 0;
}