汉子编码比字母编码长_字母/博客作者编码问题（使用动态编程）

汉子编码比字母编码长

Problem statement:

问题陈述：

Shivang is a blog writer and he is working on two websites simultaneously. He has to write two types of blogs which are:

Shivang是一位博客作家，他同时在两个网站上工作。他必须写两种类型的博客：

Technical blogs for website_1: X of this type can be written in an hour. (X will be the input)
website_1的技术博客 ：此类X可以在一小时内编写。 ( X将作为输入)
Fun blogs for website_2: Y of this type can be written in an hour. (Y will be input)
网站_2的趣味博客 ：可以在一个小时内编写此类Y。 (将输入Y )

You are to help him to save time. Given N number of total blogs, print the minimum number of hours he needs to put for combination of both the blogs, so that no time is wasted.

您是要帮助他节省时间。给定总数为N的博客，请打印他将两个博客组合在一起所需的最少小时数 ，以免浪费时间。

    Input:
N: number of total blogs
X: number of Technical blogs for website_1 per hour
Y: number of Fun blogs for website_2 per hour
Output:
Print the minimum number of hours Shivang needs to write the 
combination of both the blogs (total N). 
If it is NOT possible, print "−1".

Example:

例：

    Input:
N: 33
X: 12
Y: 10
Output:
-1 
Input:
N: 36
X: 10
Y: 2
Output:
6 (10*3+2*3)

Explanation:

说明：

    For the first case,
No combination is possible that's why output is -1.
For the second test case,
Possible combinations are 30 technical blogs (3 hours) + 6 fun blogs (3 hours)
20 technical blogs (2 hours) + 16 fun blogs (8 hours)
10 technical blogs (1 hours) + 26 fun blogs (13 hours)
0 technical blogs (0 hours) + 36 fun blogs (18 hours)
So, best combination is the first one which takes total 6 hours

The problem is basically solving equation,

问题基本上是解决方程式，

aX + bY = N where we need to find the valid integer coefficients of X and Y. Return a+b if there exists such else return -1.

aX + bY = N ，我们需要找到X和Y的有效整数系数。如果存在则返回a + b ，否则返回-1 。

We can find a recursive function for the same too,

我们也可以找到相同的递归函数，

Say,

说，

    f(n) = minimum hours for n problems
f(n) = min(f(n-x) + f(n-y)) if f(n-x), f(n-y) is solved already

We can convert the above recursion to DP.

我们可以将上述递归转换为DP。

Solution Approach:

解决方法：

Converting the recursion into DP:

将递归转换为DP：

    1)  Create DP[n] to store sub problem results
2)  Initiate the DP with -1 except DP[0], DP[0]=0
3)  Now in this step we would compute values for DP[i]
4)  for i=1 to n
if i-x>=0 && we already have solution for i-x,i.e.,DP[i-x]!=-1 
DP[i]=DP[i-x]+1;
end if
if i-y>=0 && we already have solution for i-y,i.e.,DP[i-y]!=-1)
if DP[i]!=-1
DP[i]=min(DP[i],DP[i-y]+1);
else
DP[i]=DP[i-y]+1;
End if
End if
End for
5)  Return DP[n]

C++ Implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
int minimumHour(int n, int x, int y)
{
int a[n + 1];
a[0] = 0;
for (int i = 1; i <= n; i++)
a[i] = -1;
for (int i = 1; i <= n; i++) {
if (i - x >= 0 && a[i - x] != -1) {
a[i] = a[i - x] + 1;
}
if (i - y >= 0 && a[i - y] != -1) {
if (a[i] != -1)
a[i] = min(a[i], a[i - y] + 1);
else
a[i] = a[i - y] + 1;
}
}
return a[n];
}
int main()
{
int n, x, y;
cout << "Enter total number of blogs, N:\n";
cin >> n;
cout << "Enter number of techical blogs, X:\n";
cin >> x;
cout << "Enter number of fun blogs, Y:\n";
cin >> y;
cout << "Minimum hour to be dedicated: " << minimumHour(n, x, y) << endl;
return 0;
}

Output

输出量

RUN 1:
Enter total number of blogs, N:
36
Enter number of techical blogs, X:
10
Enter number of fun blogs, Y:
2
Minimum hour to be dedicated: 6
RUN 2:
Enter total number of blogs, N:
33
Enter number of techical blogs, X:
12
Enter number of fun blogs, Y:
10
Minimum hour to be dedicated: -1