A. Potion-making

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A. Potion-making

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 ee+w⋅100 % (without rounding) magic essence and we+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
inputCopy
3
3
100
25
outputCopy
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.

题目大意

按照百分比配置药水，k为药，为水，每次向锅里加水只能加1L，问，要配置该比例的药水，最少加几次

水题

求最大公因数即可

代码

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int k;
		cin>>k;
		int w = 100 - k;
		if(k==0||w==0)
		{
			cout<<1<<endl;
		}
		else
		{
			int ans = __gcd(k,w);
			int kk = k/ans;
			int ww = w/ans;
			int a = kk + ww;
			cout<<a<<endl;
		}
	}
	return 0;
}

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