POJ 3040 Allowance

Description
As a reward for record milk production, Farmer John has decided to start paying Bessie the cow a small weekly allowance. FJ has a set of coins in N (1 <= N <= 20) different denominations, where each denomination of coin evenly divides the next-larger denomination (e.g., 1 cent coins, 5 cent coins, 10 cent coins, and 50 cent coins).Using the given set of coins, he would like to pay Bessie at least some given amount of money C (1 <= C <= 100,000,000) every week.Please help him ompute the maximum number of weeks he can pay Bessie.

解题报告:
很无聊的题，贪心一下就好，原则是选择能超过C的最小搭配，那么显然对于本身价值就大于C的直接用掉就好，如果小于的话我们选择最小搭配
考虑实现:
我们要先用掉面额比较大的，拿它去搭配较小面额，所以我们能用多少就用多少，对于剩余的我们就用小面额的补上，然后找到循环节，一次性减掉即可，这样一次至少用掉一种面额，所以复杂度\(O(N^2)\)

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define RG register
#define il inline
#define iter iterator
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
const int N=25;
struct node{
	int v,w;
	bool operator <(const node &pp)const{
		return v<pp.v;
	}
}a[N];
ll ans=0;int r[N];
void work()
{
	int n,c,lim=1e6;
	scanf("%d%d",&n,&c);
	for(int i=1;i<=n;i++){
		scanf("%d%d",&a[i].v,&a[i].w);
		if(a[i].v>=c){
			ans+=a[i].w;
			a[i].w=0;
		}
	}
	sort(a+1,a+n+1);
	int tot=0,tmp;
	while(tot<=0){
		tot=c;
		for(int i=n;i>=1;i--){
			tmp=Min(a[i].w,tot/a[i].v);
			tot-=tmp*a[i].v;
			r[i]=tmp;
		}
		if(tot){
			for(int i=1;i<=n;i++){
				if(a[i].w>r[i])tot-=a[i].v,r[i]++;
				if(tot<=0)break;
			}
		}
		if(tot>0)break;
		tmp=lim;
		for(int i=1;i<=n;i++)
			if(r[i]>0)tmp=Min(tmp,a[i].w/r[i]);
		for(int i=1;i<=n;i++){
			if(r[i]>0)a[i].w-=tmp*r[i];
			r[i]=0;
		}
		ans+=tmp;
	}
	printf("%lld\n",ans);
}

int main()
{
	work();
	return 0;
}