Dynamic Programming?

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 
     
     
     
     

 

题解：

把每一个数dp值赋初值为当前的值，每次向后加一位，最后选取最小值。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define inf  9999999
using namespace std;
int dp[1005][1005];
int e[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",&e[i]);
		}
	    memset(dp,inf,sizeof(dp));
	    for(int i=1;i<=n;i++){
	    	dp[i][i]=e[i];
	    }
		for(int i=1;i<=n;i++){
			for(int j=i+1;j<=n;j++){
				dp[i][j]=dp[i][j-1]+e[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;
}