题解1

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

这个题也是dp的题就是让你算出一个连续的最长绝对值序列，就是从第一个开始枚举，这里有个小问题，就是在dp数列中要保留原来的值，而不是abs的值，否者后面会出现问题 dp[i][j]=dp[i][j-1]+a[j]

```cpp
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int dp[1005][1005];
int a[1005];
int main()
{
	int t,k=0;
	cin>>t;
	while(t--)
    {
		k++;
		int n;
		cin>>n;
		for(int i=1;i<=n;i++)
			scanf("%d",&a[i]);
	    for(int i=1;i<=n;i++)
	    	dp[i][i]=a[i];
		for(int i=1;i<=n;i++)
			for(int j=i+1;j<=n;j++)
				dp[i][j]=dp[i][j-1]+a[j];
		int mixn=99999999;
		for(int i=1;i<=n;i++)
			for(int j=i;j<=n;j++)
				if(mixn>abs(dp[i][j]))
				    mixn=abs(dp[i][j]);
		printf("Case %d: %d\n",k,mixn);
	}
	return 0;
}