codeforces 1151 A

一个让我爆零的水题，，，，，

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!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

 1 #include <iostream>
 2 #include <cmath>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <string>
 6 #include <map>
 7 #include <iomanip>
 8 #include <algorithm>
 9 #include <queue>
10 #include <stack>
11 #include <set>
12 #include <vector>
13 //const int maxn = 1e5+5;
14 #define ll long long
15 //#define MAX INT_MAX
16 #define FOR(i,a,b) for( int i = a;i <= b;++i)
17 //#define MOD 142857
18 using namespace std;
19 char c;
20 int n,minn=INT_MAX,ans1,ans2,ans3,ans4,sum,a,b,ans;
21 string s;
22 int main()
23 {
24 //    freopen("D:\common_text\code_stream\in.txt","r",stdin);
25 //    freopen("D:\common_text\code_stream\out.txt","w",stdout);
26     cin>>n>>s;
27     for(int i=0;i<=n-4;++i)
28     {
29         sum=0;
30         int a=abs(s[i]-'A');
31         int b=abs(s[i+1]-'C');
32         int c=abs(s[i+2]-'T');
33         int d=abs(s[i+3]-'G');
34         sum+=min(a,26-a)+min(b,26-b)+min(c,26-c)+min(d,26-d);
35         if(minn>sum)
36             minn=sum;
37     }
38     cout<<minn;
39 }