CodeForces - 1525A Potion-making

Posted 2021-06-28 海岛Blog

A. Potion-making
time limit per test: 1 second
memory limit per test: 256 megabytes

You have an initially empty cauldron, and you want to brew a potion in it. The potion consists of two ingredients: magic essence and water. The potion you want to brew should contain exactly k % magic essence and (100−k) % water.

In one step, you can pour either one liter of magic essence or one liter of water into the cauldron. What is the minimum number of steps to brew a potion? You don’t care about the total volume of the potion, only about the ratio between magic essence and water in it.

A small reminder: if you pour e liters of essence and w liters of water (e+w>0) into the cauldron, then it contains $\frac{e}{e+w}$⋅100 % (without rounding) magic essence and $\frac{w}{e+w}$⋅100 % water.

Input
The first line contains the single t (1≤t≤100) — the number of test cases.

The first and only line of each test case contains a single integer k (1≤k≤100) — the percentage of essence in a good potion.

Output
For each test case, print the minimum number of steps to brew a good potion. It can be proved that it’s always possible to achieve it in a finite number of steps.

Example
input
3
3
100
25
output
100
1
4
Note
In the first test case, you should pour 3 liters of magic essence and 97 liters of water into the cauldron to get a potion with 3 % of magic essence.

In the second test case, you can pour only 1 liter of essence to get a potion with 100 % of magic essence.

In the third test case, you can pour 1 liter of magic essence and 3 liters of water.

问题链接：CodeForces - 1525A Potion-making
问题简述：（略）
问题分析：（略）
AC的C++语言程序如下：

/* CodeForces - 1525A Potion-making */

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int t, n;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);

        printf("%d\\n", 100 / __gcd(100, n));
    }

    return 0;
}