Luba And The Ticket
Luba has a ticket consisting of 6 digits. In one move she can choose digit in any position and replace it with arbitrary digit. She wants to know the minimum number of digits she needs to replace in
order to make the ticket lucky.
The ticket is considered lucky if the sum of first three digits equals to the sum of last three digits.
Input
You are given a string consisting of 6 characters (all characters are digits from 0 to 9) — this string denotes Luba's ticket. The ticket can start with the digit 0.
Output
Print one number — the minimum possible number of digits Luba needs to replace to make the ticket lucky.
Example
Input
000000
Output
0
Input
123456
Output
2
Input
111000
Output
1
Note
In the first example the ticket is already lucky, so the answer is 0.
In the second example Luba can replace 4 and 5 with zeroes, and the ticket will become lucky. It's easy to see that at least two replacements are required.
In the third example Luba can replace any zero with 3. It's easy to see that at least one replacement is required.
题意：六个数字改变最少的数字使得前三位的和等于后三位的和。
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int n,n2;
string s,ss;
int ans,ans1;
void dfs(int y,int k)
{
if((ss[3]-'0')+(ss[4]-'0')+(ss[5]-'0')==(ss[0]-'0')+(ss[1]-'0')+(ss[2]-'0'))
{
ans=min(y,ans);
}
if(y>6||k>=6)
return;
for(int i=0; i<=9; i++)
{
if(i!=s[k]-'0')
{
ss[k]=i+'0';
dfs(y+1,k+1);
ss[k]=s[k];
}
else
{
dfs(y,k+1);
}
}
return ;
}
int main()
{
cin>>s;
ss=s;
for(int i=0; i<3; i++)
{
n+=(s[i]-'0');
n2+=(s[i+3]-'0');
}
ans=100;
ans1=100;
dfs(0,0);
cout<<min(ans1,ans)<<endl;
}