Good morning [字符串]

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给定一个只有小写字母构成的非空字符串，可以从字符串中任选字符并任意规定顺序，每个字符只能用一次。

最多可以构成多少个”goodmorning”子串(可以重叠)。

比如说：字符串aaavbbbddgggooooooddmmrrnnnnii，可以构成goodmorningoodmorning，共2个。

输入
有多组测试数据，请处理到文件结束。

每组数据给定一个只有小写字母构成的非空字符串str。后台所有数据保证1 <= |str| <= 10^5。

输出
每组数据输出一个整数，表示最多可以构成的”goodmorning”子串。

样例输入
aaavbbbddgggooooooddmmrrnnnnii
goodmorninn
goodmorning
样例输出
2
0
1

#include<cstdio>#include<cstring>#include<cmath>#include<stack>#include<queue>#include<algorithm>using namespace std;#define INF 0x3f3f3f3f#define ll long longchar s[100005];int main(){    while (~scanf("%s",s)){        int len = strlen(s);        int a[150];        memset(a,0,sizeof(a));        for (int i = 0 ; i < len; ++i){            if (s[i] == 'g') a[s[i]-'a']++;            if (s[i] == 'o') a[s[i]-'a']++;            if (s[i] == 'm') a[s[i]-'a']++;            if (s[i] == 'd') a[s[i]-'a']++;             if (s[i] == 'r') a[s[i]-'a']++;             if (s[i] == 'n') a[s[i]-'a']++;             if (s[i] == 'i') a[s[i]-'a']++;            }            if (a['g'-'a'] >= 3) a['g'-'a'] += 1;            a['g'-'a']/=2;            a['o'-'a']/=3;            a['n'-'a']/=2;          int min = 0x3f3f3f3f;        if (min > a['g'-'a']) min = a['g'-'a'];        if (min > a['o'-'a']) min = a['o'-'a'];        if (min > a['d'-'a']) min = a['d'-'a'];        if (min > a['m'-'a']) min = a['m'-'a'];        if (min > a['r'-'a']) min = a['r'-'a'];        if (min > a['n'-'a']) min = a['n'-'a'];        if (min > a['i'-'a']) min = a['i'-'a'];        printf("%d\n",min);    }    return 0;}

一道水题 code速度太慢代码而且麻烦
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