D 4acm/ICPC Asian contest Mudanjiang Station D and K

Known notation Time Limit: 2 seconds memory limit: 131072 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

# Include <iostream> # include <cstdio> # include <cstring> # include <algorithm> using namespace STD; int Mian () {# ifdef xxz freopen ("in ", "r", stdin); # endif // xxz int t; CIN> T; while (t --) {string STR; CIN> STR; int Len = Str. length (); int num = 0, star = 0; For (INT I = 0; I <Len; I ++) {If (STR [I] = '*') Star ++; else num ++;} int ans = 0; int left_num = 0; If (Num <= Star) {left_num + = Star-num + 1; ans + = left_num;} For (INT I = 0, P = len-1; I <Len; I ++) {While (I <P & STR [p] = '*') p --; If (STR [I] = '*') {left_num --; // For example, 11×, -- and then it can be equivalent to an operation. 1*1 = 1. Now only this operation is performed on if (left_num <= 0) {swap (STR [I], STR [p]); ans ++; p --; left_num + = 2 ;}} else left_num ++ ;} cout <ans <Endl;} return 0 ;}

Domination Time Limit: 8 seconds memory limit: 131072 kb Special Judge

Edward is the headmaster of marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative chessboardNRows andMColumns.

Every day after work, Edward will place a chess piece on a random empty cell. A few days later, he found the chessboard wasDominatedBy the chess pieces. That means there is at least one chess piece in every row. Also, there is at least one chess piece in every column.

"That's interesting! "Edward said. He wants to know the expectation Number of days to make an empty chessboardN×MDominated. Please write a program to help him.

Input

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

There are only two integersNAndM(1 <=N,M<= 50 ).

Output

For each test case, output the expectation number of days.

Any solution with a relative or absolute error of at most 10-8 will be accepted.

Sample Input

21 32 2

Sample output

3.0000000000002.666666666667

The first probability of slag DP ....

Expected idea: calculate the probability first, and then use the expected formula to calculate the number of stones in each row, set DP [I] [J] [k] To show that J exists after I put a stone, and K in the column has the probability of having at least one stone. Then we will discuss the four cases. 1. add 1, 2 to both rows and columns. rows plus 1, 3. add 14. rows and columns are not added. Note that when I = N and J = m, no DP [I] [J] [k] cannot be added. DP [I] [J] [k-1] * (I * j-k + 1) /(n * m-k + 1), because if it reaches DP [N] [m] [K-1], the game ends, it is impossible to reach DP [N] [m] [k] from this.

#include <iostream>#include <cstdio>#include <cstring>using namespace std;double dp[60][60][2510];int main(){#ifdef xxz    freopen("in","r",stdin);#endif // xxz    int T;    cin>>T;    while(T--)    {        int n,m;        cin>>n>>m;        memset(dp,0,sizeof(dp));        dp[0][0][0] = 1.0;        for(int  i = 1; i <= n ; i++)            for(int j = 1; j <= m; j++)                for(int k = 1; k <= n*m; k++)                {                    double temp = n*m - k + 1;                    if(i == n && j == m)                    {                        dp[i][j][k] = dp[i - 1][j - 1][k - 1] *(1.0*(n-i+1)*(m - j+1) / temp)                                      + dp[i-1][j][k-1]*(1.0*(n - i+1)*j/temp)                                      +dp[i][j-1][k-1]*(1.0*i*(m - j+1 )/temp);                    }                    else                    {                        dp[i][j][k] = dp[i - 1][j - 1][k - 1] *(1.0*(n-i+1)*(m - j+1) / temp)                                      + dp[i-1][j][k-1]*(1.0*(n - i+1)*j/temp)                                      +dp[i][j-1][k-1]*(1.0*i*(m - j+1 )/temp)                        +dp[i][j][k-1]*(1.0*(i*j - k+1)/temp);                    }                }        double ans = 0;        for(int i = 1; i <= n*m; i++)        {            ans += dp[n][m][i] * i;        }        printf("%.12lf\n",ans);    }    return 0;}

D 4acm/ICPC Asian contest Mudanjiang Station D and K