hdu 4104 Discount (思维 数学归纳法）

Discount

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1072    Accepted Submission(s): 645

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

   
   
   
   

    
    
    
    4
1 2 3 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    11
   
   
   
   

 

Source

2011 Alibaba-Cup Campus Contest

 

Recommend

lcy

思路：

先将a[i]排序，假设前i个元素，已经能表示1~sum之间的数了，那么a[i+1]加进来之后，

1.如果a[i+1]>sum+1，那么sum+1肯定表示不了，ans=sum+1。

2.如果a[i+1]<=sum+1，那么1~sum+a[i+1]就都能表示了，产生新的sum，进行下一循环。

ps：懂了之后就是满简单的一道思维题了，可是比赛时就是想不到，思维能力太烂了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
//#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define mod 1000000007
#define INF 0x3f3f3f3f
using namespace std;

typedef long long ll;
int n,m,ans,cnt,flag,sum;
int a[maxn];
int dp[maxn];

void solve()
{
    int i,j,u,v;
    sum=0;
    for(i=1;i<=n;i++)
    {
        if(a[i]>sum+1)
        {
            ans=sum+1;
            return ;
        }
        sum+=a[i];
    }
    ans=sum+1;
}
int main()
{
    int i,j;
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a+1,a+n+1);
        solve();
        printf("%d\n",ans);
    }
    return 0;
}