codeforeces 845B

题解

codefores 845B

原题

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.

题目大意

给你一个长度为6的字符串，可以使任意数变为另一个数，问最多需要几次变换使前三个数的和等于后三个数的和。

样例

simple1

000000

0

simple2

123456

2

simple3

111000

1

思路

变换的情况只有4种，简单判断一下（变一次有一种，变两次有三种）。

1. 要使变换次数最少尽量减少后三个中最大的
2. 或是补上前三个中最小的（默认前三个的和小于后三个的）

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
char g[8];
int sum1,sum2,a[3],b[3];
void Swap(int a[3],int b[3])
{
    for(int i=0;i<3;i++) swap(a[i],b[i]);
}  

int main()
{
    while(gets(g)!=NULL)
    {
        for(int i=0;i<3;i++){
            a[i]=g[i]-'0';
            sum1+=g[i]-'0';
        }
        for(int i=3;i<6;i++){
            b[i-3]=g[i]-'0';
            sum2+=g[i]-'0';
        }
        sort(a,a+3);
        sort(b,b+3);
        if(sum1>sum2){
            Swap(a,b);
            swap(sum1,sum2);
        }
        int p=sum2-sum1;
        if(!p)
            printf("0\n");
        else if(p<=9-a[0]||p<=b[2])//只有一个数字变换的最大情况
            printf("1\n");
        else if(p<=9-a[0]+9-a[1]||p<=b[2]+b[1]||p<=9-a[0]+b[2])//变换两个数字变换有三种情况 两前，两后，一前一后；
            printf("2\n");
        else            //两种情况都不满足 一定是第三种
            printf("3\n");
    }
    return 0;
}