Multiple knapsack, simplicity and binary optimization

#include<bits/stdc++.h>
using namespace std;
/*int main()
{
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;i++)
	{
		cin>>v[i]>>w[i]>>s[i];
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<=m;j++)
		{
			for(int k=0;k<=s[i]&&k*v[i]<=j;k++)
			{
				dp[i][j]=max(dp[i][j],dp[i-1][j-k*v[i]]+k*w[i]);
			}
		}
	}
	cout<<dp[n][m];
	return 0;
	
}*/
/*
	 good , Now optimize :
	 First , We use full backpacks   To optimize , because max It's impossible to do subtraction .( It's too late ）
	 OK, now let's start binary optimization ：
	 hypothesis i Yes n individual , Now let's put this n Divide into groups , The number of each group is 
	0,1,2,4,8,...,2^(logn-1)( among logn Round down ),c;
	 among 1,2,4,8,... ,2^(logn-1) The sum of is less than or equal to n, If it is less than ( also 
	 That is to say n No 2 The whole power of c=n-sum( 1,2,4,8,... ,2^(logn-1));
	 And through the thought of multiplication  0,1,2,4,8,...,2^(logn-1),c These numbers , Yes. 
	 take 0~n Every number of is expressed . So when we separate all kinds of objects ,
	 We got many groups , Each group has its own volume 、 value . Therefore, the problem of multiple backpacks 
	 Transformed into 01 The problem with backpacks .
	 Code implementation : 

*/
const int N = 2e5+ 10;
int dp[N];
int v[N],w[N],cnt=0;
int main()
{
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;i++)
	{
		int a,b,c,k=1;
		cin>>a>>b>>c;
		while(k<=c)
		{
			v[++cnt]=a*k;
			w[cnt]=b*k;
			c-=k;
			k*=2;
		}
		if(c>0)
		{
			v[++cnt]=a*c;
			w[cnt]=b*c;
		}
	}
	for(int i=1;i<=cnt;i++)
	{
		for(int j=m;j>=v[i];j--)
		{
			dp[j]=max(dp[j],dp[j-v[i]]+w[i]);
		}
	}
	cout<<dp[m];
	return 0;
}