贪心+模拟 ZOJ 3829 Known Notation

题目传送门

 /*
     题意：一串字符串，问要最少操作数使得成为合法的后缀表达式
     贪心+模拟：数字个数 >= *个数+1 所以若数字少了先补上在前面，然后把不合法的*和最后的数字交换，记录次数
             岛娘的代码实在难懂啊~
 */
 /************************************************
 * Author        :Running_Time
 * Created Time  :2015-8-16 14:29:49
 * File Name     :K.cpp
  ************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = 1e3 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 char str[MAXN];

 int main(void)    {     //ZOJ 3829 Known Notation
     int T;  scanf ("%d", &T);
     while (T--) {
         scanf ("%s", str);
         int x = , len = strlen (str);
         for (int i=; i<len; ++i)  {
             if (str[i] == '*')  x++;
         }
         if (x == ) {
             puts (""); continue;
         }
         int ans = max (x +  - (len - x), );   int n = ans;
         for (int i=; i<len; ++i)   {
             if (str[i] == '*')  {
                 if (n <= ) n++, ans++;
                 else    n--;
             }
             else    n++;
         }
         printf ("%d\n", ans);
     }

     return ;
 }