【题解】LuoGu1776：宝物筛选

原题传送门
暴力dp
$dp_{i,j}=max(dp_{i-1,j-w*k}+v*k)$
可以用倍增思想， 把物品用二进制分组
比如 $9->1,2,4,2$
这样在复杂度里面可以把一个 $m$优化成 $logm$
复杂度满足

Code：

#include <bits/stdc++.h>
#define maxn 110
#define maxm 40010
using namespace std;
struct node{
	int v, w;
}a[maxm];
int n, W, dp[maxm], tot;

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

int main(){
	n = read(), W = read();
	for (int i = 1; i <= n; ++i){
		int v = read(), w = read(), m = read(), x = 1;
		while (x <= m) a[++tot].v = x * v, a[tot].w = x * w, m -= x, x <<= 1;
		if (m) a[++tot].v = m * v, a[tot].w = m * w;
	}
	for (int i = 1; i <= tot; ++i)
		for (int j = W; j >= a[i].w; --j) dp[j] = max(dp[j], dp[j - a[i].w] + a[i].v);
	printf("%d\n", dp[W]);
	return 0;
}