【P1018 乘积最大非完全题解】DP部分

最新推荐文章于 2022-02-17 20:20:59 发布

原创最新推荐文章于 2022-02-17 20:20:59 发布 · 172 阅读

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博客介绍了如何利用动态规划（DP）解决P1018题目，该题目需要通过切m刀使区间l到r的乘积最大化。博主指出由于涉及高精度计算，直接求解较为复杂，故借鉴了前一题目的思路进行调整。

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P1018
这个题要得100分是需要高精度的太懒了
所以借着上题的思路改一下因为上题是分成m部分这题是是切m刀
大同小异 dp[i][l][r] 代表区间 l 到 r 切了 i 刀

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
ll dp[7][45][45],arr[45],a[45][45];
char str[45];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,m;scanf("%d%d",&n,&m);
    scanf("%s",str+1);
    for(int i = 1;i<=n;++i) arr[i] = str[i]-'0',a[i][i] = arr[i];
    for(int i = 1;i<n;++i)
        for(int j = i+1;j<=n;++j) a[i][j] = a[i][j-1]*10+a[j][j];
    for(int i = 1;i<=n;++i) dp[0][1][i] = a[1][i];
    for(int i = 1;i<=m;++i)
        for(int l = 1;l<=n;++l)
            for(int r = l + i - 1;r<=n;++r)
                for(int k = l+i-2;k<r;++k)
                    {
                        dp[i][l][r] = max(dp[i][l][r],dp[i-1][l][k]*a[k+1][r]);
                    }
    printf("%lld\n",dp[m][1][n]);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}