P1018 学习笔记

省流：高精度，但不是 python

题目传送门

经典 DP。

我们设 \(dp_{i,j}\) 表示前 \(i\) 位用 \(j\) 个乘号的最大值。

边界：\(dp_{i,0}=a_{1,i}\)

转移方程可以考虑对于 \((i,j)\) 枚举乘号的位置 \(k\) 进行转移，于是：

\[dp_{i,j}=\max_{k=j}^{i-1} dp_{k,j-1} \cdot a_{k+1,i} \]

答案就是 \(dp_{n,m}\)。

但是——这题要装高精度！！

你可以写封装的，我就直接字符串了。

code

#include <bits/stdc++.h>
#define DEBUG
#define Ofile(s) freopen(s".in", "r", stdin), freopen (s".out", "w", stdout)
#define Cfile(s) fclose(stdin), fclose(stdout)
#define fast ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
using namespace std;

using ll = long long;
using ull = unsigned long long;
using lb = long double;

constexpr int mod = 998244353;
constexpr int maxn = 45;
constexpr int maxk = 1005;

int n, k;
string s;

string dp[maxn][maxn], a[maxn][maxn];

string init(int l, int r){
	int pos = l;
	while (pos < r && s[pos] == '0')
		pos++;
	return s.substr(pos, r - pos + 1);
}

string max(string x, string y){
	if (x.empty())
		return y;
	if (y.empty())
		return x;
	if (x.size() != y.size())
		if (x.size() > y.size())
			return x;
		else 
			return y;
	else 
		if (x > y)
			return x;
		else 
			return y;
}

string mul(string x, string y){
	int lenx = x.size(), leny = y.size(), lenz = x.size() + y.size();
	string z("");
	if (x == "0" || y == "0")
		return "0";
	int p[maxk], q[maxk], r[maxk];
	memset(p, 0, sizeof(p));
	memset(q, 0, sizeof(q));
	memset(r, 0, sizeof(r));
	for (int i = 0; i < lenx; i++)
		p[lenx - i] = x[i] - '0';
	for (int i = 0; i < leny; i++)
		q[leny - i] = y[i] - '0';
	for (int i = 1; i <= lenx; i++)
		for (int j = 1; j <= leny; j++)
			r[i + j - 1] += p[i] * q[j];
	for (int i = 1; i <= lenz; i++)
		r[i + 1] += r[i] / 10, r[i] %= 10;
	if (!r[lenz])
		--lenz;
	for (int i = lenz; i >= 1; i--)
		z += (char)(r[i] + '0');
	return z;
}

int main() {
	fast;
	cin >> n >> k >> s;
	s = " " + s;
	for (int i = 1; i <= n; i++)
		for (int j = i; j <= n; j++)
			a[i][j] = init(i, j);
	for (int i = 1; i <= n; i++)
		dp[i][0] = a[1][i];
	for (int j = 1; j <= k; j++)
		for (int i = j + 1; i <= n; i++)
			for (int t = j; t < i; t++)
				dp[i][j] = max(dp[i][j], mul(dp[t][j - 1], a[t + 1][i]));
	cout << dp[n][k];
	return 0;
}