Known Notation ZOJ

Do you know reverse Polish notation (RPN)? It is a known notation in the area of mathematics and computer science. It is also known as postfix notation since every operator in an expression follows all of its operands. Bob is a student in Marjar University. He is learning RPN recent days.

To clarify the syntax of RPN for those who haven't learnt it before, we will offer some examples here. For instance, to add 3 and 4, one would write "3 4 +" rather than "3 + 4". If there are multiple operations, the operator is given immediately after its second operand. The arithmetic expression written "3 - 4 + 5" in conventional notation would be written "3 4 - 5 +" in RPN: 4 is first subtracted from 3, and then 5 added to it. Another infix expression "5 + ((1 + 2) × 4) - 3" can be written down like this in RPN: "5 1 2 + 4 × + 3 -". An advantage of RPN is that it obviates the need for parentheses that are required by infix.

In this problem, we will use the asterisk "*" as the only operator and digits from "1" to "9" (without "0") as components of operands.

You are given an expression in reverse Polish notation. Unfortunately, all space characters are missing. That means the expression are concatenated into several long numeric sequence which are separated by asterisks. So you cannot distinguish the numbers from the given string.

You task is to check whether the given string can represent a valid RPN expression. If the given string cannot represent any valid RPN, please find out the minimal number of operations to make it valid. There are two types of operation to adjust the given string:

1. Insert. You can insert a non-zero digit or an asterisk anywhere. For example, if you insert a "1" at the beginning of "2*3*4", the string becomes "12*3*4".
2. Swap. You can swap any two characters in the string. For example, if you swap the last two characters of "12*3*4", the string becomes "12*34*".

The strings "2*3*4" and "12*3*4" cannot represent any valid RPN, but the string "12*34*" can represent a valid RPN which is "1 2 * 34 *".

Input

There are multiple test cases. The first line of input contains an integer Tindicating the number of test cases. For each test case:

There is a non-empty string consists of asterisks and non-zero digits. The length of the string will not exceed 1000.

<h4< dd="">

Output

For each test case, output the minimal number of operations to make the given string able to represent a valid RPN.

<h4< dd="">

Sample Input

31*111*234***

<h4< dd="">

Sample Output

10
2
观察发现n个*消耗n+1个数字是肯定的
所以当*个数大于数字个数减一时必须补数字  因为没空格所以数字多是可以的，因为可以把多个数字看成一个
个数确定好了之后还有位置问题 因为在串中也可能出现数字
#include <iostream>#include<cstdio>#include<cstring>using namespace std;int main(){    int n;    cin>>n;    while(n--)    {        char s[1005];        getchar();        scanf("%s",s);        int num1=0,num2=0;        for(int i=0;s[i]!='\0';i++)        {            if(s[i]=='*')num1++;            else num2++;        }        int ans=num1-num2+1;        if(ans<0)ans=0;        int leftnum=ans;              //leftnum表示的是当前剩余数字个数        int numpos=strlen(s)-1;        for(int i=0;s[i]!='\0';i++)        {            while(i<numpos&&s[numpos]=='*')numpos--;   //记录最靠后的数字位置            if(s[i]=='*')            {                leftnum--;                if(leftnum-1<0)              {                swap(s[i],s[numpos]);                leftnum+=2;                numpos--;                ans++;              }            }            else leftnum++;        }        cout<<ans<<endl;    }    return 0;}

不够的情况 这时候可以把最后的数字跟*交换位置