Maximum Subsequence Sum 【DP】

最新推荐文章于 2022-07-15 14:20:26 发布

原创最新推荐文章于 2022-07-15 14:20:26 发布 · 252 阅读

2 ·

CC 4.0 BY-SA版权

文章标签：

#DP

动态规划专栏收录该内容

24 篇文章

订阅专栏

本文探讨了给定整数序列中寻找具有最大和的连续子序列的问题，并提供了使用动态规划方法解决该问题的详细步骤及示例输入输出。特别关注如何确定最大子序列的起始和结束位置。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

思路
用DP的方法动态更新答案每次对每一个数求和当和小于 0 的时候就重置为0 然后 sum > ans 的时候就更新

然后有一个难点就是要输出该子串的首尾元素

一共有几种情况

0.最大和序列中有负数
1.并列和对应i相同但是不同j，即尾部有0
2.1个正数
3.全是负数
4.负数和0
5.最大和前面有一段0
6.最大N

我们只需要标记一下起点就可以了
然后更新答案的那个就是尾部

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a) memset(a, 0, sizeof(a))

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair<string, int> psi;
typedef pair<string, string> pss;

const double PI = 3.14159265358979323846264338327;
const double E = exp(1);
const double eps = 1e-6;

const int INF = 0x3f3f3f3f;
const int maxn = 1e3 + 5;
const int MOD = 1e9 + 7;

int arr[maxn];

int main()
{
    int t;
    cin >> t;
    for (int l = 1; l <= t; l++)
    {
        int n, k;
        scanf("%d%d", &n, &k);
        ll ans = 0;
        for (int i = 0; i < n; i++)
            scanf("%d", &arr[i]);
        sort(arr, arr + n);
        int len = n - k + 1;
        for (int i = 0; i < len; i++)
            ans -= arr[i];
        for (int i = n - 1, j = 0; j < len; i--, j++)
            ans += arr[i];
        printf("Case #%d: %d\n", l, ans);
    }
}