Hdu 4223 Dynamic Programming?

Dynamic Programming?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2188    Accepted Submission(s): 1005

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

#include"stdio.h"
#include"string.h"
#define N 1101

int abs(int a)
{
	if(a>0)return a;
	return -a;
}

int main()
{
	int T;
	int n;
	int A[N];
	int dp[N];
	int i,j,t;
	int ans,cnt;

	scanf("%d",&T);
	cnt=1;
	while(T--)
	{
		scanf("%d",&n);
		scanf("%d",&A[0]);
		ans=dp[0]=abs(A[0]);
		for(i=1;i<n;i++)
		{
			scanf("%d",&A[i]);
			dp[i]=abs(A[i]);
			t=A[i];
			//因为要求是连续的，所以得从i-1开始
			for(j=i-1;j>=0;j--)
			{
				t+=A[j];
				if(abs(t)<dp[i])
					dp[i]=abs(t);
			}
			if(dp[i]<ans)ans=dp[i];
		}
		printf("Case %d: %d
",cnt++,ans);
	}
	return 0;
}