Dynamic Programming?

2183 단어 dp
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

 
문제 풀이:
매 수의 dp값을 현재 값으로 설정하고, 매번 뒤로 한 자리를 추가하고, 마지막으로 최소 값을 선택합니다.
#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
",k,mixn); } return 0; }

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