Time Limit: 1 Sec Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/540/problem/ADescription
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; }