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。
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
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;
}
ACM Rounders 题解
178

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