ZOJ - 2781 Rounders_zojrounders-优快云博客

本文链接：https://blog.youkuaiyun.com/acm_1361677193/article/details/47080817

本文详细解读了一个数值四舍五入的编程挑战，包括输入解析、四舍五入逻辑实现及输出展示。通过实例代码，深入探讨了如何根据不同数量级进行精确的数值调整。

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Rounders

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %lld & %llu

Submit Status

Description

Introduction

For a given number, if greater than ten, round it to the nearest ten, then (if that result is greater than 100) take the result and round it to the nearest hundred, then (if that result is greater than 1000) take that number and round it to the nearest thousand, and so on ...

Input

Input to this problem will begin with a line containing a single integer n indicating the number of integers to round. The next n lines each contain a single integer x (0 <= x <= 99999999).

Output

For each integer in the input, display the rounded integer on its own line.
Note: Round up on fives.

Sample Input

Sample Output

Source

South Central USA Regional Programming Contest 2006

分析：

水题。

ac代码：

#include <iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    int n,i,x,x2;
    int c;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d",&x);
        c=0;
        x2=x;
        for(;x2;x2/=10)
        {
            c++;
        }
        //printf("c=%d\n",c);
        for(i=1;i<c;i++)
        {
            if(x%(int)(pow(10,i))>=(int)5*(pow(10,i-1)))
            {
                x=x-x%(int)(pow(10,i))+(int)(pow(10,i));
                //printf("xx=%d",x);
            }
            else
            {
                x=x-x%(int)(pow(10,i));
               //printf("xx=%d",x);
            }
        }
        printf("%d\n",x);
    }
    return 0;
}