hiho_1061_beautiful_string

题目大意

    题目连接：beautiful string 
    写代码之前，考虑清楚流程，以及需要维护的变量....

实现

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#include<iostream> #include<stdio.h> #include<string.h> #include<stack> #include<vector> #include<unordered_set> #include<unordered_map> using namespace std; char str[10 * 1024 * 1024 + 2]; int main(){     int T;     scanf("%d", &T);     while (T--){         int n;         scanf("%d", &n);         getchar();         scanf("%s", str);               int i = 0;         bool valid = false;         int beg = 0, end = 0;         int count_1 = 0; //前2个字符连续的个数         int count_2 = 0; //前1个字符连续的个数         while (beg < n && !valid){             char cur = str[beg]; //当前字符             while (end < n && !valid){                 if (str[end] == cur){                                       //如果前两个字符都有，且当前字符数目大于等于中间字符数目，                     //则肯定可以形成一个valid的子串                     if (count_2 && count_1 && end - beg + 1 >= count_2){                         valid = true;                         break;                     }                     end++;                 }                 else{                     if (str[end] - cur != 1){//出现了一个不连续字符，则清空count_1,count_2                         count_1 = 0;                         count_2 = 0;                     }                     else{                                               //当前字符之前的那个字符的连续个数大于count_2的个数，                         //则只能以当前字符作为新的中间字符 count_2，且                         //count_1 清空                         if (count_2 && count_2 < end - beg){                             count_1 = 0;                             count_2 = end - beg;                         }                         else{                             //更新count_1 为count_2                             //更新 count_2为 end-beg;                             count_1 = count_2;                             count_2 = end - beg;                         }                                           }                     break;                 }             }             beg = end;         }         if (valid)             printf("YES\n");         else             printf("NO\n");     }     return 0; }