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1007 Maximum Subsequence Sum

Statement

Metadata

  • 作者: CHEN, Yue
  • 单位: 浙江大学
  • 代码长度限制: 16 KB
  • 时间限制: 200 ms
  • 内存限制: 64 MB

Given a sequence of K integers { N_1, N_2, …, N_K }. A continuous subsequence is defined to be { N_i, N_{i+1}, …, N_j } where 1 \le i \le j \le 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 (\le 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

Solution

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

#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 = 1e4 + 5;
const int MOD = 1e9 + 7;

int arr[maxn];

int main() {
    int n;
    cin >> n;
    int l, r, ans = -1, num, temp = 0, vis, flag = 0;
    for (int i = 0; i < n; i++) {
        scanf("%d", &arr[i]);
        if (i == 0 || flag) {
            vis = arr[i];
            flag = 0;
        }
        temp += arr[i];
        if (temp < 0) {
            temp = 0;
            flag = 1;
            continue;
        }
        if (temp > ans) {
            ans = temp;
            l = vis;
            r = arr[i];
        }
    }
    if (ans != -1)
        printf("%d %d %d\n", ans, l, r);
    else
        printf("%d %d %d\n", 0, arr[0], arr[n - 1]);
}
Last update: May 4, 2022
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