大意:给定后缀表达式, 每次操作可以添加一个字符, 可以交换两个字符的位置, 相邻数字可以看做一个整体也可以分开看, 求合法所需最少操作数.
数字个数一定为星号个数+1, 添加星号一定不会更优.
先判断若星号过多, 直接在最左边添上数字, 遍历过程中若星号还多的话把星号与右侧数字交换.
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endifint n;
char s[N];void work() {scanf("%s", s+1);n = strlen(s+1);int star = 0, num = 0;REP(i,1,n) {if (s[i]=='*') ++star;else ++num;}int left_num = 0, ans = 0;if (num<=star) {left_num += star-num+1;ans += left_num;}int now = n;REP(i,1,n) {while (i<now&&s[now]=='*') --now;if (s[i]=='*') {if (--left_num<1) {++ans,--now;left_num+=2;}}else ++left_num;}printf("%d\n", ans);
}int main() {int t;scanf("%d", &t);while (t--) work();
}
