[luoguP3092] [USACO13NOV]没有找零No Change（状压DP + 二分）

传送门

先通过二分预处理出来，每个硬币在每个商品处最多能往后买多少个商品

直接状压DP即可

f[i]就为，所有比状态i少一个硬币j的状态所能达到的最远距离，在加上硬币j在当前位置所能达到的距离，所有的取max

是满足最优解性质的

#include <cstdio>
#include <iostream>
#include <algorithm>
#define N 17
#define max(x, y) ((x) > (y) ? (x) : (y))

int n, k, s1, s2, ans = -1;
int a[N], sum[100001], c[N][100001], f[1 << N];

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

int main()
{
	int i, j;
	k = read();
	n = read();
	for(i = 1; i <= k; i++) a[i] = read(), s1 += a[i];
	for(i = 1; i <= n; i++) sum[i] = read() + sum[i - 1];
	for(i = 1; i <= k; i++)
		for(j = 1; j <= n; j++)
			c[i][j] = std::upper_bound(sum + j, sum + n + 1, a[i] + sum[j - 1]) - sum - 1;
	for(i = 1; i < (1 << k); i++)
		for(j = 1; j <= k; j++)
			if(i & (1 << j - 1))
				f[i] = max(f[i], c[j][f[i ^ (1 << j - 1)] + 1]);
	for(i = 1; i < (1 << k); i++)
		if(f[i] == n)
		{
			s2 = 0;
			for(j = 1; j <= k; j++)
				if(i & (1 << j - 1))
					s2 += a[j];
			ans = max(ans, s1 - s2);
		}	
	printf("%d
", ans);
	return 0;
}