1007. Maximum Subsequence Sum (25)-PAT甲级真题(最大连续子序列和、动态规划dp)

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


题目理解

给出一个序列,求子序列中sum最大的子序列
输出最大自序列和 子序列第一个元素和最后一个元素
如果不唯一,输出id最小的那组
如果所有数为负数,sum最大为0,输出首位元素

分析

  • sum为最大和,temp为临时最大和,leftid和rightid为子序列首尾下标,tempid记录leftid的临时下标
  • temp=temp+v[i],当temp>sum时,更新sum,leftid,rightid
  • 当temp<0,无论来什么值都应该舍弃temp<0前的值,因为负数只会拉低sum的值,不会增加sum
  • 舍弃后temp=0,更新tempid
  • 输出化sum=-1,leftid=0, rightid=n-1,若最后sum仍然小于<0,说明所有数字都小于0,输出0 0 n-1
#include 
#include 

using namespace std;

int main() {
    int n;
    cin >> n;
    vector v(n);
    int leftid = 0, rightid = n - 1, tempid = 0, sum = -1, temp = 0;
    for (int i = 0; i < n; i++) {
        cin >> v[i];
        temp += v[i];
        if (temp < 0) {
            temp = 0;
            tempid = i + 1;
        } else if (temp > sum) {
            sum = temp;
            leftid = tempid;
            rightid = i;
        }
    }
    if (sum < 0)
        sum = 0;
    cout << sum << " " << v[leftid] << " " << v[rightid] << endl;
    return 0;
}

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