poj3040 发工资（贪心）

题目传送门

题目大意：给一个人发工资，给出不同数量不同面额，（大面额一定是小面额的倍数），问最多能发几天，（每天实发工资>=应发工资）。

思路：首先，将大于等于c的面额的钱直接每个星期给奶牛一张，将面额大于等于c的前去除。然后从大到小开始选择，要选择的金额尽可能的接近c，如果刚好能够凑足c就作为当前的一种方案，如果不能凑足c那就再从小的开始选，要选出一种的总额不少于c但尽量接近c作为当前的方案，然后计算如果按照如此方案最多可以给奶牛多少周，然后按照相同的方法再选方案，一直选到选择的金额不能凑足c为止。

为什么这样做是对的呢？

题目中说大面值一定是小面值的倍数，所以，如果一个数字可以用大面值凑，那必定可以用小面值凑，也就是说，小面值的功能比大面值强。我们在能选择大面值的时候，尽量选择大面值。在这个过程中我们一直保证m<=c,但如果我们不能使m<=c，即即便我们用了最小的面额也会超值，那就只能浪费超过的这一部分了，并且此时浪费的是最少的。（如果先选小的再选大的，浪费的肯定大于等于此时的情况）

代码。

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
using namespace std;
typedef long long ll;
struct dian {
	int val,to;
} a[30];
bool cmp(dian a,dian b) {
	return a.val<b.val;
}
int make[30];
int main() {
	int n,c;
	cin>>n>>c;
	for(int i=1; i<=n; i++) {
		cin>>a[i].val>>a[i].to;
	}
	sort(a+1,a+1+n,cmp);
	int ans=0;
	for(int i=n; i>0; i--) {
		if(a[i].val>=c) {
			ans+=a[i].to;
			n--;
		} else break;
	}
	bool flag;
	while(1) {
		flag=false;
		memset(make,0,sizeof(make));//make表示  某一种货币在一种工资组合中的使用次数
		int tepc=c;
		int i;
			for(i=n; i>0; i--) {
				if(a[i].to) {
					int m,num;
					num=tepc/a[i].val;
					m=min(num,a[i].to);//尽可能的多用，但不能超过总数
					tepc-=m*a[i].val;
					make[i]=m;
					if(tepc==0) { //此处只会>=0
						flag=true;
						break;
					}
				}
			}
			if(tepc>0) {
				for(i=1; i<=n; i++) {
					if(a[i].to>make[i]) {//make里面存的是当前这一轮用过的   除了用过的  还有剩下的  才可以用这种货币
						while(make[i]<a[i].to) {
							tepc-=a[i].val;
							make[i]++;
							if(tepc<0) {//此处不会=0 因为=0的情况上面已经排除掉了
								flag=true;
								break;
							}
						}
					}
					if(flag)break;
				}
			}
		if (!flag) break;
			int cnt=0x3f3f3f3f;
			for(int i=1;i<=n;i++){
				if(make[i])
				cnt=min(cnt,a[i].to/make[i]);//最小的组合
			}
			ans+=cnt;
			for(int i=1;i<=n;i++){
				a[i].to-=make[i]*cnt;
			}
	}
	cout<<ans<<endl;
}

Allowance

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5167		Accepted: 2032

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.

Input

* Line 1: Two space-separated integers: N and C 

* Lines 2..N+1: Each line corresponds to a denomination of coin and contains two integers: the value V (1 <= V <= 100,000,000) of the denomination, and the number of coins B (1 <= B <= 1,000,000) of this denomation in Farmer John's possession.

Output

* Line 1: A single integer that is the number of weeks Farmer John can pay Bessie at least C allowance

Sample Input

3 6
10 1
1 100
5 120

Sample Output

111

Hint

INPUT DETAILS: 
FJ would like to pay Bessie 6 cents per week. He has 100 1-cent coins,120 5-cent coins, and 1 10-cent coin. 

OUTPUT DETAILS: 
FJ can overpay Bessie with the one 10-cent coin for 1 week, then pay Bessie two 5-cent coins for 10 weeks and then pay Bessie one 1-cent coin and one 5-cent coin for 100 weeks.