DP 01 knapsack problem

Konjac water first hair blog, is a template title tortured afternoon (laughs)

Luo Gu P1060

code show as below:

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int chengji[100000000], jiage[30], jiazhi[30];
int n, m;//n为钱数，m为个数 
int main(void){
cin>>n>>m;
for(int i=0; i<m; i++){
cin>>jiage[i]>>jiazhi[i];
jiazhi[i]=jiage[i]*jiazhi[i];
}
for(int i=0 ; I <m; I ++ ) {
 for ( int J = n-; J> = [I] jiage; J, ) { 
Chengji [J] = max (Chengji [J], Chengji [J-jiage [I]] + jiazhi [I]); 
} 
} 
COUT << Chengji [n-];
 return  0 ; 
}

This is a one-dimensional array optimized for ease of understanding, paste a two-dimensional array (Valley from Los solution to a problem)

#include <iostream>

using namespace std;

int f[30][100000];
int w[10000];
int v[10000];

int main()
{
    int n,m;
    int i,j,k;
    cin>>m>>n;
    //提前相乘
    for(i=1;i<=n;i++)
    {
        cin>>w[i]>>v[i];
        v[i]*=w[i];
    }
    for(int i=1; I <= n-; I ++ ) 
    { 
    // most critical location backpack 01, in order to prevent repeatedly applied in the same article, the need to search backwards, 01 which is completely different from the Backpacks backpack 
        for ( int C = 0 ; C < = m; C ++ ) 
        { 
            F [I] [C] = F [I- . 1 ] [C];
             IF (C> = W [I]) 
            F [I] [C] = max (F [I] [C ], F [I- . 1 ] [CW [I]] + V [I]); 
        } 
    } 
    COUT << F [n-] [m];
     return  0 ; 
}

One-dimensional array and a two-dimensional array are the two ways of the article when the last bit number i (1 <= i <= m) how can the product takes the maximum value, except that a different starting point.

Thinking in the two-dimensional array of sequence number i in the object is the last one, calculating the remaining volume of j (0 <= j <= n) can take when the maximum product to,

Thinking one-dimensional array is assumed that the i-th items into the bag, how to get the remainder can take maximum.

I want to make a start with a recursive method, but can not find expression backpack space restrictions, so do strike.

Established a 2020 flag, put the problem in a recursive method to do it again XD