Codeforces

来源：互联网发布：linux informix 目录编辑：程序博客网时间：2024/05/16 11:03

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.

Examples
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：给出6个数字，求换多少次才能使前三个和等于后三个和。

解题思路：水题，这里给出搜索的做法……之前做搜索比较多，看到什么都想搜。实际上这题有更简单的做法，可以直接模拟。

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <algorithm>#include <string.h>#include <math.h>#include <vector>#include <set>#include <map>#include <queue>#include <stack>#define ll long longusing namespace std;string str;vector<int> a;map<string,int> vis;struct point{    string a;    int step;    point(string aa,int b){        a=aa;        step=b;    }};int bfs(){    queue<point> que;    que.push(point(str,0));    vis[str]=1;    while(!que.empty()){        point tp=que.front();        que.pop();        int fs=tp.a[0]-'0'+tp.a[1]-'0'+tp.a[2]-'0';        int bs=tp.a[3]-'0'+tp.a[4]-'0'+tp.a[5]-'0';        if(fs==bs)            return tp.step;        for(int i=0;i<7;i++)//枚举替换6个数字            for(int j=0;j<10;j++){//枚举替换成9种                string temp=tp.a;                temp[i]='0'+j;                if(vis[temp]==0)//如果没被访问过                    que.push(point(temp,tp.step+1));            }    }}int main(){    cin>>str;    cout<<bfs()<<endl;    return 0;}

阅读全文

0 0