Discount(HDU 4104,求数列和子集问题）

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本文探讨了一种算法，用于解决商家提供的折扣策略下，顾客无法达到理想购物预算的问题。通过输入商品价格列表，算法计算出无法实现的最小预算。实例分析揭示了通过排序和累加商品价格来寻找最小不可达预算的策略。

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/*http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22157#problem/D
D - Discount
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit

Status

Practice

Description

All the shops use discount to attract customers, but some shops doesn’t give direct discount on their goods, instead, they give discount only when you bought more than a certain amount of goods. Assume a shop offers a 20% off if your bill is more than 100 yuan, and with more than 500 yuan, you can get a 40% off. After you have chosen a good of 400 yuan, the best suggestion for you is to take something else to reach 500 yuan and get the 40% off.
For the customers’ convenience, the shops often offer some low-price and useful items just for reaching such a condition. But there are still many customers complain that they can’t reach exactly the budget they want. So, the manager wants to know, with the items they offer, what is the minimum budget that cannot be reached. In addition, although the items are very useful, no one wants to buy the same thing twice.

Input
The input consists several testcases.
The first line contains one integer N (1 <= N <= 1000), the number of items available.
The second line contains N integers Pi (0 <= Pi <= 10000), represent the ith item’s price.

Output
Print one integer, the minimum budget that cannot be reached.

Sample Input
4
1 2 3 4

Sample Output
11
题意：这道题卡就卡在题意理解上了，看了半天还是没有把题意弄懂，看了挺多人AC，就应该知道这题不难，事实证明很简单
总的意思就是给你一列数列，根据其和的子集，求出一个最小的数不在其子集中；例如：输入
3
3 4 7
则有其和的结果：3
4
7
3+7=10；
4+7=11；
3+4+7=14
故结果为1；
其实不难发现，如果数列中最小的如果大于1的话，最终结果自然就是1(因为无论怎么做，始终得不到最小值为1的预算)
*/

i#nclude<stdio.h>
#include<algorithm>
#include<string.h>
#include <iostream>
using namespace std;
const int maxn=1000+10;
int p[maxn];
int main()
{int i,n,sum;
 while(scanf("%d",&n)!=EOF)
  {


           for(i=0;i<n;i++)
           {scanf("%d",&p[i]);
           }
           sort(p,p+n);
           sum=0;
           for(i=0;i<n;i++)
           {
           if(sum<(p[i]-1))//这说明当前的sum至少比p[i]小2，得不到sum+1 ，退出
             break;
             sum+=p[i];
           }
           printf("%d\n",sum+1);
}
return 0;


}