POJ 3040Allowance(贪心好题目)

Allowance

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1039		Accepted: 444

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

如上面的实例FJ有n种面额的硬币，每种硬币的面额排序之后相邻两个之间整数倍关系一样。每种硬币的数量有nod[i].num个，每周都需要付给佣人至少money金币，只可以>=，不能少于这个数目。问最多可以给多少周。

首先：

1.nod[i].val>=money，即面额大于等于money，则直接取它的数目即可。

2。依次从大面额到小面额试着取，如果刚好面额之和=money，那就更好了，这个面额可以用小的凑起来，所以选大的合适。

3.如果大的凑不出来相等的，那么选最大的加起来且<money的，然后从小的开始往大的选，如果能选到的话，就直接选择。

因为是整数倍关系，至少是2倍的关系，比如a[1]=1,a[2]=2,a[3]=4,a[4]=8,那么每次都是a[i]比a[1]~a[i-1]所有面额之和还大，所以保证了第三种策略的正确性。

题目地址：Allowance

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

int res;
int need[25];

struct node
{
    int val;
    int num;
}nod[25];

int cmp(node p1,node p2)
{
    if(p1.val>=p2.val) return 1;
    return 0;
}

int main()
{
    int n,money,i,j,t,tmp;
    while(cin>>n>>money)
    {
        for(i=0;i<n;i++)
            scanf("%d%d",&nod[i].val,&nod[i].num);
        sort(nod,nod+n,cmp);  //面额从大到小排序

        int res=0;
        for(i=0;i<n;i++)      //第一阶段，取>=money面额的
        {
            if(nod[i].val<money) break;
            res+=nod[i].num;
        }
        t=i;

        while(1)
        {
            memset(need,0,sizeof(need));
            int rest=money;         //还需要多少
            for(i=t;i<n;i++)    //尽可能用大面值凑够,如果刚好放下就用大面额，不够用小面额凑
            {
                if(!nod[i].num||!rest)
                    continue;
                tmp=rest/nod[i].val;
                tmp=min(nod[i].num,tmp);
                need[i]=tmp;
                rest-=need[i]*nod[i].val;
            }

            if(rest)    //需要小面额凑
            {
                for(i=n-1;i>=t;i--)
                    if(nod[i].val>=rest&&(nod[i].num-need[i]))
                    {
                        need[i]++;
                        rest=0;
                        break;
                    }
                if(rest)    break;
            }

            int ans=1e9;
            for(i=t;i<n;i++)
                if(need[i])
                    ans=min(ans,nod[i].num/need[i]);   //最多放多少组一样的
            res+=ans;

            for(i=t;i<n;i++)
                nod[i].num-=ans*need[i];
        }
        cout<<res<<endl;
    }
    return 0;
}

/*
3 50
40 1
20 1
10 4

4 42
40 1
20 1
10 6
5 3
*/