hdoj Discount 4104 （简单数学归纳法&思想）

Discount

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1947    Accepted Submission(s): 1129

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
 
//题意：
给你n中钞票面值，每种钞票只有一张，问你他们不能表示的最小价值是多少。
用数学归纳法，从小到大来找用他们的和来遍历他们能表示的数，若1---x都能被表示，但x+1不能被表示，则最小值为x+1.
#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>using namespace std;int a[1010];int sum[1010];int main(){int n,i;while(scanf("%d",&n)!=EOF){memset(a,0,sizeof(a));memset(sum,0,sizeof(sum));for(i=1;i<=n;i++)scanf("%d",&a[i]);sort(a+1,a+n+1);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]+1)break;}printf("%d\n",sum[i-1]+1);}return 0;}