dp--2019南昌網絡賽B-Match Stick Game

Xiao Ming recently indulges in match stick game and he thinks he is good at it. His friend Xiao Jun decides to test him. Xiao Jun gives him an expression of length , made by match sticks and asks him to calculate the maximum value of the expression by moving any match sticks (but he can’t discard any of them). The expression is made up of some numbers, plus signs and minus signs represented as A_1 ?op_1 ?A_2 ?op_2 ?A_3 ?op_3 ?\cdots A_{m - 1} ?op_{m - 1} ?A_mA

1. The new expression should also be made up of mm
   numbers and m - 1m?1 operators.
2. The number of digits per number should keep consistent with the original.
3. There couldn’t be any leading zeros per number.

Input

The first line consists of a single integer TT denoting the number of test cases.

There’re two lines in each test case.

The first line contains an integer nn

A string of length nn follows in the next line, denoting the expression given.

The expression is guaranteed to be valid.

Output

For each test case, print a single integer denoting the maximum result of the expression.

Constraints

\[ 1≤n≤100 \]

Note

Expression with the maximum result for the second sample is 7 - 17?1 .

Expression with the maximum result for the second sample is 7 + 7 + 97+7+9.

樣例輸入復制

3
3
1-1
3
1+1
5
1+2+3

樣例輸出復制

0
6
23

題意

給你一條式子，式子有火柴棒組成，可以移動火柴棒，要求：式子中運算符號的數目不變，即進行運算的數字數量不變。每組進行運算的數的位數不變。火柴棒的數目不變。式子最後得到的結果最大。

思路

把式子看成多組數進行加減運算

預處理:mx[i][j]表示在一組數中，第i位用了j根火柴所能達到的最大值，mi[i][j]同理為最小值

考慮加減號，狀態轉移看代碼，dp[i][j]表示這條式子中用i組數，j根火柴所能達到的最大值

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <sstream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <iomanip>
#include <stack>

using namespace std;

typedef long long LL;
const int INF = 0x3f3f3f3f;
const int N = 1e7 + 50;
const int MOD = 1e9 + 9;

#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define F(i, l, r) for(int i = l;i <= (r);++i)
#define RF(i, l, r) for(int i = l;i >= (r);--i)

int p[] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};//0123456789
int sum, num, n;
string s;
LL dp[105][1005], dig[105];//第i組數第用j根火柴能到的最大值，每組數的位數
LL mx[15][1005], mi[15][1005];//一組數中第i位用j火柴可以到的最值

void solve()
{
    fill(mx[0], mx[0] + 15 * 1005, -1);
    fill(mi[0], mi[0] + 15 * 1005, INF);
    mx[0][0] = mi[0][0] = 0;
    F(i, 1, 11)//根據題目，最多有9位數
        F(j, 0, i * 7)//一個數字最多用7根火柴
            F(k, 0, 9)
            {
                if(p[k] > j) continue;
                mx[i][j] = max(mx[i][j], mx[i - 1][j - p[k]] * 10 + k);
                mi[i][j] = min(mi[i][j], mi[i - 1][j - p[k]] * 10 + k);
            }

    memset(dp, -1, sizeof(dp));
    memset(dig, 0, sizeof(dig));
    int len = s.size();
    sum = 0, num = 1;//火柴數，組數
    F(i, 0, len - 1)
    {
        if(s[i] == '+') {num++; sum += 2;}
        else if(s[i] == '-') {num++; sum++;}
        else {sum += p[s[i] - '0']; dig[num]++;}
    }
    F(i, 1, sum)
        dp[1][i] = mx[dig[1]][i];
    F(i, 2, num)//num組數
        F(j, 0, sum)//一共的火柴數
            F(k, 1, 7 * dig[i])//該組數所用的火柴數
            {
                if(j >= 2 + k && dp[i - 1][j - 2 - k] != -1 && mx[dig[i]][k] != -1)//火柴數目夠，且前一組數有答案，且這一組數用k根火柴有最值
                    dp[i][j] = max(dp[i][j], dp[i - 1][j - 2 - k] + mx[dig[i]][k]);
                if(j >= 1 + k && dp[i - 1][j - 1 - k] != -1 && mi[dig[i]][k] != INF)
                    dp[i][j] = max(dp[i][j], dp[i - 1][j - 1 - k] - mi[dig[i]][k]);
            }
}

int main()
{
    int T;
    cin >> T;
    while(T--)
    {
        cin >> n >> s;
        solve();
        cout << dp[num][sum] << endl;
    }
    return 0;
}

參考博客

dp--2019南昌網絡賽B-Match Stick Game