AtCoder Beginner Contest 192 F - Potion 背包dp

传送门

题意： 给你$n$个数，让后让你选出来$k$个$A$，把他们求和，之后再递增$k$直到正好达到$x$，求最小的递增次数。

思路： 转化一下题意就是求$\sum A=x(\mathrm{m}\mathrm{o}\mathrm{d}len)$，且$\sum A$最大，考虑如何解决$\sum A$最大的问题。
设$f[i][j][k]$表示前$i$个数选了$j$个且$\mathrm{m}\mathrm{o}\mathrm{d}len$为$k$，那么转移就比较明显了：$$f[i][j][k]=max(f[i-1][j][k],f[i-1][j-1][(k+len-(a[i]\mathrm{m}\mathrm{o}\mathrm{d}len))\mathrm{m}\mathrm{o}\mathrm{d}len])$$答案就是$max(ans,(x-f[n][len][x\mathrm{m}\mathrm{o}\mathrm{d}len])/len)$。

**//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;

//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;

const int N=210,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;

int n;
LL x,ans=1e18,a[N];
LL f[N][N][N];// 前i个数选j个%len余数为k

int main()
{
    
    
//	ios::sync_with_stdio(false);
//	cin.tie(0);

    scanf("%d%lld",&n,&x);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    for(int len=1;len<=n;len++)
    {
    
    
        memset(f,-INF,sizeof(f));
        f[0][0][0]=0;
        for(int i=1;i<=n;i++)
        {
    
    
            f[i][0][0]=0;
            for(int j=1;j<=len;j++)
                for(int k=0;k<len;k++)
                    f[i][j][k]=max(f[i-1][j][k],f[i-1][j-1][(k+len-(a[i]%len))%len]+a[i]);
        }
        if(f[n][len][x%len]>=0) ans=min(ans,(x-f[n][len][x%len])/len);
    }
    printf("%lld\n",ans);


	return 0;
}
/*

*/