Zoj 3829 known notation

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

Question and code:

This was a bit difficult during the competition. This question is to give a suffix expression to see if it is a reasonable expression. If it is unreasonable, we have two operations: 1. Any two characters are exchanged. 2. add a number or *. It is actually a greedy question. We will find that if there are N x numbers, there must be at least N + 1 numbers in the expression. Otherwise, we need to add numbers to satisfy n + 1. The greedy strategy is to put the numbers at the beginning of the expression, and then when a * is determined, if the number of the preceding numbers is less than 2, we will find a number next to this * and change it to another position until all the numbers are calculated.

#include <iostream>#include <cstdio>#include <cstring>using namespace std;char s[1010];int main(){    int cas;    scanf("%d",&cas);    while(cas--)    {        scanf("%s",s);        int ans=0,len=strlen(s),p=-1,nstar=0,nnum=0;        for(int i=0;i<len;i++)        if(s[i]=='*')        {            if(p<0) p=i;            nstar++;        }        else nnum++;        if(p==-1)        {            printf("0\n");            continue;        }        int k=len-1,num=0;        if(nstar>=nnum)        {            ans=(nstar-nnum+1);            num=ans;        }        if(s[len-1]!='*')        {            ans++;            swap(s[p],s[len-1]);        }        for(int i=0;i<len;i++)        if(s[i]=='*')        {            if(num>=2) num--;            else            {                ans++;                while(k>i)                {                    if(s[k]!='*')                    {                        swap(s[i],s[k]);                        num++;                        break;                    }                    k--;                }            }        }        else num++;        printf("%d\n",ans);    }    return 0;}

Zoj 3829 known notation