[BZOJ1531] [POI2005] Bank notes [多重背包]

[ $\frak{Link}$]

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把朴素多重背包的式子稍作转化，得到：
$\mathcal{f_{j}=max\{f_{k'}-k'value\;|\;k'\in[j'-count_i,j']\}+j'value,\;j'cost_i+Remainder=j}$（枚举 $\mathcal{Remainder}$）
看了半天感觉不太对结果发现这道题也是代价等于价值，，

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#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cstring>
using namespace std;
int n, m;
int a[205];
int c[205];
pair<int, int> q[20005];
int qh, qt;
int f[20005];
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &a[i]);
	}
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &c[i]);
	}
	scanf("%d", &m);
	memset(f, 0x3f, sizeof(f));
	f[0] = 0;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j < a[i]; ++j) {
			qh = 1;
			qt = 0;
			for (int x, k = 0; ; ++k) {
				if ((x = k * a[i] + j) > m) break;
				while (qh <= qt && q[qh].second < k - c[i]) {
					++qh;
				}
				while (qh <= qt && q[qt].first >= f[x] - k) {
					--qt;
				}
				q[++qt] = make_pair(f[x] - k, k);
				f[x] = min(f[x], q[qh].first + k);
			}
		}
	}
	printf("%d", f[m]);
	return 0;
}