【P1043 数字游戏】 dp

最新推荐文章于 2023-10-20 19:29:50 发布

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本文详细解析了一道关于区间划分的算法题，通过动态规划（DP）算法求解最大和最小得分。讨论了如何利用前缀和进行优化，以及在实现过程中遇到的常见错误，如除法操作符的误用。

P1043
PS：做这题心态崩了
一开始不知道原来是按顺序分以为折半二进制枚举
然后发现是按顺序分
dp[i][l][r] 代表区间 l 到区间 r 内分了 i 段的答案
那么枚举 i , l , r
用中间值 k 去分割区间那么就是左边部分 dp[i-1][l][k] * 右边的前缀和预处理
那么因为前面有了i-1段所以右端点必须从左端点 + i - 1开始
然后又开始因为求前缀和除法忘记加括号 de了半小时BUG
心态崩了

/*
    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);
int sum[105],dp[10][105][105],dp_[10][105][105],arr[105];

int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,m,ans = 0x3f3f3f3f,ans_ = 0;scanf("%d%d",&n,&m);
    for(int i = 1;i<=n;++i) scanf("%d",&arr[i]),arr[i+n] = arr[i];
    for(int i = 1;i<=2*n;++i) sum[i] = sum[i-1] + arr[i];
    memset(dp_,0x3f,sizeof(dp_));
    for(int i = 1;i<=2*n;++i)
        for(int j = i;j<=2*n;++j)
            dp[1][i][j] = dp_[1][i][j] = ((sum[j]-sum[i-1])%10+10)%10;
    for(int i = 2;i<=m;++i)
        for(int l = 1;l<=2*n;++l)
            for(int r = i + l - 1;r<=2*n;++r)
                for(int k = i+l-2;k<r;k++)
                    {
                        dp[i][l][r] = max(dp[i][l][r],dp[i-1][l][k] * (((sum[r]-sum[k])%10+10)%10));
                        dp_[i][l][r] = min(dp_[i][l][r],dp_[i-1][l][k] * (((sum[r]-sum[k])%10+10)%10));
                    }
    for(int i = 1;i<=n;++i)
    {
        ans_ = max(dp[m][i][i+n-1],ans_);
        ans = min(dp_[m][i][i+n-1],ans);
    }
    printf("%d\n%d\n",ans,ans_);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}