https://codeforces.com/contest/71/problem/B
以上是本题地址，大家可以注册，然后在上面练习。
/*
A progress bar 进度条 is an element of graphical interface that displays the progress of a process for this very moment before it is completed.
Lets take a look at the following form of such a bar.
A bar is represented as squares 正方形, located in line.
To add clarity 为了更加清晰, let's number them with positive integers from 1 to z from the left to the right.
Each square has saturation 饱和度 (ai for the i-th square),
which is measured 度量 by an integer from 0 to k.
When the bar for some i is displayed,
squares 1,2,..., i- 1 has the saturation k,
squares i+1,i+2, ..., n has the saturation 0,
and the saturation of the square i can have any value from 0 to k.
So some first squares of the progress bar always have the saturation.
Some last squares always have the saturation 0.
And there is no more than one square that has the saturation different from 0 and k.
The degree of the process's completion is measured in percents.
Let the process be t% completed. Then the following inequation is fulfilled:
An example of such a bar can be seen on the picture.
n = 10， k = 10， t = 54
a: 10 10 10 10 10 4 0 0 0 0
For the given n, k, t determine the measures of saturation for all the squares ai of the progress bar.
Examples
Input
10 10 54
Output
10 10 10 10 10 4 0 0 0 0
Input
11 13 37
Output
13 13 13 13 0 0 0 0 0 0 0
*/
原题
分析：10，10，54 就显示5个10，1个4，其余0
公式 5 = (n*k * t%) / 整除 k
公式 4 = (n*k * t%) % 取余 k
从以上就可以知道，这个程序大概意思。
以下是本人写的代码（这次未参考别人），由于简单，所以一次成功！
Codeforces Beta Round 65 (Div. 2)
cpp
/*
A progress bar is an element of graphical interface that displays the progress of a process
for this very moment before it is completed.
Lets take a look at the following form of such a bar.
A bar is represented as squares, located in line.
To add clarity, let's number them with positive integers from 1 to z from the lett to the right.
Each square has saturation (a; for the i-th square),
which is measured by an integer from 0 to k.
When the bar for some i is displayed,
squares 1,2,..., i- 1 has the saturation k,
squares i+1,i+2, ..., n has the saturation 0,
and the saturation of the square i can have any value from 0 to k.
So some first squares of the progress bar always have the saturation.
Some last squares always have the saturation 0.
And there is no more than one square that has the saturation different from 0 and k.
The degree of the process's completion is measured in percents.
Let the process be t% completed. Then the following inequation is fulfilled:
An example of such a bar can be seen on the picture.
n = 10 k = 10 t = 54
a: 10 10 10 10 10 4 0 0 0 0
For the given n, k, t determine the measures of saturation for
all the squares ai of the progress bar.
Examples
Input
10 10 54
Output
10 10 10 10 10 4 0 0 0 0
Input
11 13 37
Output
13 13 13 13 0 0 0 0 0 0 0
*/
#include <iostream>
using namespace std;
void cal_progress_bar(int n, int k, int t)
{
int product = n*k;
int c1 = 0;
c1 = t*product*0.01;
int number1 = 0;
int number2 = 0;
number1 = c1 / k;
number2 = c1 % k;
char space1 = ' ';
for (int i = 0; i < number1; i++)
{
cout << k << space1;
}
int len = 0;
if (number2>0)
{
cout << number2 << space1;
len= n - number1 - 1;
}
else
{
len = n - number1;
}
for (int i = 0; i < len; i++)
{
cout << 0 << space1 ;
}
}
int main()
{
int n = 0, k = 0, t = 0;
while (cin >> n >> k >> t)
{
cal_progress_bar(n, k, t);
}
return 0;
}