Zoj 3829 known notation (2014 Mudanjiang H)

Question link: http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemid = 5383

Known notation Time Limit: 2 seconds memory limit: 65536 KB

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 learned it before, we will offer some examples here. for instance, to add 3 and 4, one wocould 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 wocould 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 integerTIndicating 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.

Output

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

Sample Input

31*111*234***

Sample output

102

Author: Chen, Cong
Source: The 2014 ACM-ICPC Asia Mudanjiang Regional Contest
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I have been busy with various things recently and haven't written a blog for a long time...

This is the Peony H question. At that time, I had been entangled in a probability DP... I didn't think deeply about this question until the competition was over...

First, we can complete the string first, because the number must be '*' + 1. After that, we found that for '*', we

In fact, you only need to leave it behind as much as possible .. Then scan it from start to end.

But remember to be careful in some special situations ..

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<algorithm>#include<string>#include<map>#include<set>#include<vector>#include<bitset>#include<cstdlib>#define CLR(A) memset(A,0,sizeof(A))using namespace std;int main(){    int T;    while(~scanf("%d",&T)){        while(T--){            string str;            cin>>str;            int cnum=0,cop=0,ans=0;            for(int i=0;str[i];i++){                if(str[i]>='0'&&str[i]<='9') cnum++;                else cop++;            }            if(cnum<cop+1) {                ans+=(cop+1-cnum);                cnum=cop+1-cnum;            }            else cnum=0;            if(cop==0){                cout<<"0"<<endl;                continue;            }            int len=str.size();            int last=len-1;            bool flag=0;            while(last>=0&&!(str[last]>='0'&&str[last]<='9')) last--;            for(int i=0;str[i];i++){                if(i>=last) break;                if(str[i]>='0'&&str[i]<='9'){ cnum++;}                else{                    if(cnum<2){                        ans++;                        cnum++;                        last--;                        flag=1;                        while(last>=0&&!(str[last]>='0'&&str[last]<='9')) last--;                    }                    else cnum--;                }            }            if(flag==0&&str[len-1]!='*') ans++;            cout<<ans<<endl;        }    }    return 0;}

Zoj 3829 known notation (2014 Mudanjiang H)