codeforces 1151 A

一个让我爆零的水题，，，，，

codeforces 1151A   1000分

题意：一个字符串，单个字符可以一步可以变成左右两个（Z可以变成Y，A），问最低多少步可以产生“ACTG”；

错因：错误的理解了s[i]-'0'的含义了，0的ASCLL码是48!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

 #include <iostream>
 #include <cmath>
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <map>
 #include <iomanip>
 #include <algorithm>
 #include <queue>
 #include <stack>
 #include <set>
 #include <vector>
 //const int maxn = 1e5+5;
 #define ll long long
 //#define MAX INT_MAX
 #define FOR(i,a,b) for( int i = a;i <= b;++i)
 //#define MOD 142857
 using namespace std;
 char c;
 int n,minn=INT_MAX,ans1,ans2,ans3,ans4,sum,a,b,ans;
 string s;
 int main()
 {
 //    freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
 //    freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
     cin>>n>>s;
     for(int i=0;i<=n-4;++i)
     {
         sum=;
         int a=abs(s[i]-'A');
         int b=abs(s[i+]-'C');
         int c=abs(s[i+]-'T');
         int d=abs(s[i+]-'G');
         sum+=min(a,-a)+min(b,-b)+min(c,-c)+min(d,-d);
         if(minn>sum)
             minn=sum;
     }
     cout<<minn;
 }