Zoj problem set-3829 known notation (Greedy)

Zoj problem set-3829 known notation (Greedy)

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question: Give You A suffix expression (only digits and symbols), but the space of this suffix expression is unfortunately lost, now I will give you a suffix expression like this, asking you at least how many operations are required to turn this expression into legal.
operation:
1. Insert'' or a number (one-digit) anywhere in the expression ).
2. Swap the characters at any two positions of the expression.

Solution:
At first, I thought it was complicated, and the result was still missing some kind of situation, so I couldn't go through it all the time; that is, when I got stuck with '', the numbers were not enough to be inserted or replaced.
In fact, just think like this: first, the number of numbers should at least be the number of symbols + 1. First, calculate the number of numbers and symbols. If a number is not enough to be inserted, it is better to replace it. If a number is inserted at the beginning, the better it is. If a number is inserted at the beginning, the better it is, the better it is.
Note: all are numbers.

Code:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1005;char str[maxn];int main () {    int T;    scanf ("%d", &T);    while (T--) {        scanf ("%s", str);        int num, op, ans;        int len = strlen (str);        ans = num = op = 0;        for (int i = 0; i < len; i++)            if (str[i] == ‘*‘)                op++;        num = len - op;        if (!op) {            printf ("0\n");            continue;        }        if (str[len - 1] != ‘*‘) {            ans++;            for (int i = 0; i < len; i++)                if (str[i] == ‘*‘) {                    swap(str[i], str[len - 1]);                    break;                }        }        int cnt = 0;        for (int i = 0; i < len; i++) {            if (str[i] == ‘*‘) {                if (cnt > 1)                    cnt--;                else {                    if (num >= op + 1) {                        for (int j = len - 1; j >= 0; j--)                            if (str[j] != ‘*‘) {                                swap(str[j], str[i]);                                cnt++;                                ans++;                                break;                            }                    } else {                        ans++;                        num++;                        if (!cnt) {                            i--;                            cnt = 1;                        }                    }                }            } else                cnt++;        }        printf ("%d\n", ans);    }    return 0;}

Zoj problem set-3829 known notation (Greedy)