Luba And The Ticket CodeForces - 845B

最新推荐文章于 2018-08-03 15:33:57 发布

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本文介绍了一个有趣的问题：如何通过最少的数字替换使一张由六个数字组成的票成为“幸运票”。所谓的幸运票是指票的前三个数字之和等于后三个数字之和。文中提供了一种有效的算法来解决这一问题，并通过几个实例进行了说明。

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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

Output

Input

Output

Input

Output

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.

每一位都选差最大的就可以了比如前三位比后三位小那就前三位里找9-该位差最大的然后后三位里找最大的选他们俩中最大的。

#include <iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;

int main()
{
    char s[10];
    int a[10],b[10];
    while(scanf("%s",s)!=-1)
    {
        int sum1=0,sum2=0;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i=0;i<3;i++)
            a[i]=s[i]-'0',sum1+=a[i];
        for(int i=3;i<6;i++)
            b[i-3]=s[i]-'0',sum2+=b[i-3];
        int tmp=sum1-sum2;
        //cout<<tmp<<endl;
        sort(a,a+3);
        sort(b,b+3);
        if(tmp==0)
            cout<<"0"<<endl;
        else
        {
            if(tmp<0)
            {
                tmp=-tmp;
                int ans=0;
                int i=0,j=2;
                while(tmp>0)
                {
                    int f1,f2;
                    f1=9-a[i];
                    f2=b[j];
                    if(f1>=f2)
                    {
                        tmp-=f1;
                        i++;

                    }
                    else
                        tmp-=f2,j--;
                    ans++;
                }
                printf("%d\n",ans);
            }
            else
            {

                int ans=0;
                int i=0,j=2;
                while(tmp>0)
                {
                    int f1,f2;
                    f1=9-b[i];
                    f2=a[j];
                    if(f1>=f2)
                    {
                        tmp-=f1;
                        i++;

                    }
                    else
                        tmp-=f2,j--;
                    ans++;
                }
                printf("%d\n",ans);
            }
        }

    }
    return 0;
}