hdu 4223 Dynamic Programming?

Posted 2020-08-12 xjhz

Dynamic Programming?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Dynamic Programming, short for DP, is the favorite of iSea. It is a method for solving complex problems by breaking them down into simpler sub-problems. It is applicable to problems exhibiting the properties of overlapping sub-problems which are only slightly smaller and optimal substructure.
Ok, here is the problem. Given an array with N integers, find a continuous subsequence whose sum’s absolute value is the smallest. Very typical DP problem, right?

 

 

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case includes an integer N. Then a line with N integers Ai follows.

Technical Specification
1. 1 <= T <= 100
2. 1 <= N <= 1 000
3. -100 000 <= Ai <= 100 000

 

 

Output

For each test case, output the case number first, then the smallest absolute value of sum.

 

 

Sample Input

2 2 1 -1 4 1 2 1 -2

 

 

Sample Output

Case 1: 0 Case 2: 1

 

 

Author

[email protected]

 

 

Source

首届华中区程序设计邀请赛暨第十届武汉大学程序设计大赛

思路：暴力，我以为会T，还想用treap优化一下。。。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=2e5+10,M=4e6+10,inf=1e9+10,mod=1e9+7;
const ll INF=1e18+10;
int a[N];
int pre[N];
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
           scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
            pre[i]=pre[i-1]+a[i];
        int ans=inf;
        for(int i=1;i<=n;i++)
            for(int t=i;t<=n;t++)
            {
                ans=min(ans,abs(pre[t]-pre[i-1]));
            }
        printf("Case %d: %d\n",cas++,ans);
    }
    return 0;
}