o.boj 1120 Rounders

ACM Rounders 题解
注:最近这一系列ACM的内容,都是2年多之前的代码,自己回顾一下。
 
Rounders
 
Submit: 283   Accepted:179
Time Limit: 1000MS  Memory Limit: 65536K
Description
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.

Sample Input

9
15
14
4
5
99
12345678
44444445
1445
446


Sample Output

20
10
4
5
100
10000000
50000000
2000
500


Source

2006 South Central USA Regional Programm


#include <iostream>

using namespace std;

int main()
{
    int N, carry, num, len, temp;
    
    cin >> N;
    while (N--)
    {
        carry = 0;
        len = 1;
        temp = 0;
        
        cin >> num;
        
        if (num > 10)
        {
            while (num > 10)
            {
                temp = num % 10;
                if ((temp+carry) >= 5)
                    carry = 1;
                else 
                    carry = 0;
                len ++;
                num /= 10;
            }
            cout << num + carry;
            for (int j = 0; j < len - 1; j++)
                cout << "0";
            cout << endl;
        }
        else
            cout << num << endl;        
    }
    // system("pause");
    return 0;
}


评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值