HDOJ 题目4104 Discount（数学，技巧）



Discount

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1437    Accepted Submission(s): 865

Problem 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 P_i (0 <= P_i <= 10000), represent the ith item’s price.

 

Output

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

 

Sample Input

41 2 3 4

 

Sample Output

11

 

Source

2011 Alibaba-Cup Campus Contest

 

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题目大意：给你n个数，每个数只能用一次，问不能组合的最小的数

ac代码

#include<stdio.h>#include<string.h>#include<stdlib.h>int a[10010],sum[10010];int cmp(const void *a,const void *b){return *(int *)a-*(int *)b;}int main(){int n;while(scanf("%d",&n)!=EOF){int i;for(i=1;i<=n;i++)scanf("%d",&a[i]);qsort(a+1,n,sizeof(a[1]),cmp);memset(sum,0,sizeof(sum));//sum[0]=a[0];for(i=1;i<=n;i++){sum[i]=sum[i-1]+a[i];}for(i=1;i<=n;i++){if(a[i]>sum[i]+1)break;}printf("%d\n",sum[i-1]+1);}}