Topic description:
Given three integers X, Y, Z, several operations are required to make X, Y, and Z equal. There are two operations:
1. Select two numbers from X, Y, Z and add 1 to each.
2. Choose a number from X, Y, Z and add 2.
Find the minimum number of operations required.
Topic idea:
1. Sort the three numbers X, Y, Z, here is the descending sort. Assuming that X>Y>Z after sorting, if YZ is an odd number, then if (Y - Z)/2 operations 2 are performed on Z, Z will become Y-1, that is, Y=Y at this time , Z=Y-1. Then at this time, you only need to perform XY operations 1 on Y and Z first, so that X=X, Y=X, Z=X-1. Then perform one operation 1 on X, Y, and finally perform one operation on Z 2. Finally, X=Y=Z. In this case, a total of X - Y +(Y - Z)/2 + 2 operations are required.
2. If YZ is an even number, then (Y - Z)/2 operations are performed on Z, and Z becomes Y, that is, Y = Y, Z = Y at this time. Then at this time, you only need to perform X - Y operations 1 on Y and Z. In this case, a total of X - Y +(Y - Z)/2 operations are required.
code show as below:
#include<iostream>
#include<cassert>
#include<vector>
#include<stack>
#include<cstdio>
#include<unordered_map>
#include<queue>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
class Solution {
public:
int minCount(int a[])
{
int x = a[0],y = a[1],z = a[2];
if((y - z) & 1)
{
return x - y + (y - z) / 2 +2;
}
else
{
return x - y + (y - z) / 2;
}
}
};
bool cmp(int a,int b)
{
return a > b;
}
intmain()
{
int a[]={2,9,5};//Test case
Solution s;
sort(a,a+3,cmp);
cout<<s.minCount(a)<<endl;;
}