HDOJ 1003 Max Sum(動態規劃)

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

 

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

 

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

暴力求解，會超時

#include<stdio.h>
int main() 
{
	int n;
	scanf("%d", &n);
	for (int k = 1; k <= n;k++) {
		int i, j, num=0,max=0,s,e;
		int a[30], sum[30][30] = {0};
		
		scanf("%d", &num);
		for (i = 0; i < num; i++) {
			scanf("%d", &a[i]);
		}
		for (i = 0; i < num; i++) {
			sum[i][i] = a[i];
			if (sum[i][i] > max) {
				max = sum[i][i];
				s = i + 1, e = i + 1;
			}
			for (j = i + 1; j < num; j++) {
				sum[i][j] = sum[i][j - 1]+a[j];
				if (sum[i][j] > max) {
					max = sum[i][j];
					s = i+1, e = j+1;
				}
			}
		}
		printf("Case %d：\n", k);
		printf("%d %d %d\n\n", max, s, e);
	}
	return 0;
}

正解：動態規劃法

#include <iostream>
using namespace std;

int main()
{
    int t;
    cin>>t;
    for(int i=0;i<t;i++)
    {
        int n;
        cin>>n;
        int sum = 0, max = -99999;
        int curhead=1, rear=1, head=1;
        for(int j=0;j<n;j++)
        {
            int temp;
            cin>>temp;
            if(sum<0)
            {
                curhead = j+1;
                sum = temp;
            }else
            {
                sum += temp;
            }
            if(sum>max)
            {
                rear = j + 1;
                head = curhead;
                max = sum;
            }
        }
        cout<<"Case "<<i+1<<":"<<endl;
        cout<<max<<' '<<head<<' '<<rear<<endl;
        if(i!=t-1)
            cout<<endl;
    }
}