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[];
//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=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
}
//************************************************************************************** int main()
{
string s1,s2;
int n=read();
cin>>s1>>s2;
ll ans=;
for(int i=;i<n;i++)
{
int k=s1[i]-'';
int m=s2[i]-'';
int ans1=;
int ans2=;
while(k!=m)
{
k++;
k%=;
ans1++;
}
k=s1[i]-'';
while(k!=m)
{
k--;
if(k<)
k+=;
ans2++;
}
ans+=min(ans1,ans2);
}
cout<<ans<<endl; }
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