洛谷 P2983 [USACO10FEB]购买巧克力Chocolate Buying

购买巧克力Chocolate Buying

乍一看以为是背包，然后交了一个感觉没错的背包上去。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
//Mystery_Sky
//
#define M 10000000
#define ll long long
ll f[M];
ll p[M], num[M], sum;
ll n, v;
int main() {
	scanf("%lld%lld", &n, &v);
	for(int i = 1; i <= n; i++) scanf("%lld%lld", &p[i], &num[i]);
	for(int i = 1; i <= n; i++) {
		for(ll k = 1; k <= num[i]; k++) {
			for(ll j = v; j >= p[i]; j--) {
				f[j] = max(f[j], f[j-p[i]]+1);
			}
		}
	}
	printf("%lld\n", f[v]);
	return 0;
}

结果无情30分。

---

看了一下数据范围，再仔细想了下，发现不是dp，贪心就可以了，从小到大排序费用，再从小到大买，到买不起为止即可。

---

Code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
//Mystery_Sky
//
#define ll long long
#define M 1000000
struct node {https://i.cnblogs.com/EditCategories.aspx?catid=1
	ll p, c;
}m[M];
ll ans, b;
int n;
inline bool cmp(node a, node b)
{
	return a.p < b.p;
}

int main() {
	scanf("%d %lld", &n, &b);
	for(int i = 1; i <= n; i++) scanf("%lld%lld", &m[i].p, &m[i].c);
	sort(m+1, m+1+n, cmp);
	for(int i = 1; i <= n; i++) {
		if(b / m[i].p >=  m[i].c) {
			ans += m[i].c;
			b -= m[i].p * m[i].c;
		}
		else {
			ans += b / m[i].p;
			break;
		}
	}
	printf("%lld\n", ans);
	return 0;
}