1007 Maximum Subsequence Sum

Given a sequence of K integers { N1, N2, …, NK }. A continuous subsequence is defined to be { Ni, Ni+1, …, Nj } where 1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K(≤10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:

10 -10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4

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动态规划，求最大子序列和

设dp[i]表示前i个数的子序列（一定包含a[i]）和的最大值

若dp[i-1]<0,则dp[i-1]+a[i]<a[i];即dp[i]=a[i]

若dp[i-1]≥0,则dp[i]=dp[i-1]+a[i](dp[i]一定要包含a[i])

最会求a[i]的最大值

#include<iostream>
#include<algorithm>
#include<iomanip>
#include <ctype.h>
#include<string.h>
#include<string>
#include<cstring>
#include<stdio.h>
using namespace std;

struct Sum {
    int data = 0;
    int left, right;
};
int main() {//动态规划
    int K;
    int num[10010];
    cin >> K;
    for (int i = 0; i < K; i++) {
        cin >> num[i];
    }
    Sum sum[10010];//记录动态最大
    sum[0].data = num[0];
    sum[0].left = 0;
    sum[0].right = 0;
    for (int i = 1; i < K; i++) {
        if (sum[i - 1].data < 0) {
            sum[i].data = num[i];
            sum[i].left = i;
            sum[i].right = i;
        }
        else 
        { 
            if (num[i] == 0) {
                sum[i].data = sum[i - 1].data + num[i];
                sum[i].left = sum[i - 1].left;
                sum[i].right = sum[i-1].right;
            }
            if (num[i] != 0) {
                sum[i].data = sum[i - 1].data + num[i];
                sum[i].left = sum[i - 1].left;
                sum[i].right = i;
            }
        }
    }
    Sum max=sum[1];
    //int index;
    for (int i = 0; i < K; i++) {
        if (sum[i].data > max.data) {
            max = sum[i];
        }
    }
    if (max.data < 0)cout << 0 << ' ' << num[0] << ' ' << num[K-1];
    else cout << max.data << ' ' << num[max.left] << ' ' << num[max.right];
}