Codeforces Round #301 (Div. 2) A. Combination Lock 暴力

A. Combination Lock

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/540/problem/A

Description

Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.

The combination lock is represented by n rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?

Input

The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of disks on the combination lock.

The second line contains a string of n digits — the original state of the disks.

The third line contains a string of n digits — Scrooge McDuck's combination that opens the lock.

 

Output

Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.

 

Sample Input

5
82195
64723

 

Sample Output

13

 

HINT

 

题意

给你一个锁，问你最少转多少下，到达目标态

题解：

暴力转就好了~

代码：

 

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //§ß§é§à§é¨f§³
const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

int main()
{
    string s1,s2;
    int n=read();
    cin>>s1>>s2;
    ll ans=0;
    for(int i=0;i<n;i++)
    {
        int k=s1[i]-'0';
        int m=s2[i]-'0';
        int ans1=0;
        int ans2=0;
        while(k!=m)
        {
            k++;
            k%=10;
            ans1++;
        }
        k=s1[i]-'0';
        while(k!=m)
        {
            k--;
            if(k<0)
                k+=10;
            ans2++;
        }
        ans+=min(ans1,ans2);
    }
    cout<<ans<<endl;

}