PAT:1007 Maximum Subsequence Sum (25分)

最新推荐文章于 2025-07-25 13:08:09 发布

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本文介绍了一种高效算法来解决最大子序列和问题。通过分析整数序列，找到具有最大和的连续子序列及其范围。算法使用了动态规划的思想，能够快速定位到最优解。

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≤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

感觉这道题有点像前缀和，并不需要将前缀放入数组里，用一个变量来当就行了，当变量小于0，置零，大于最大值，就更该最大值和左右边界。还是很容易理解的

#include<bits/stdc++.h>
using namespace std;
int s[10005];
int main()
{
    int n;
    scanf("%d",&n);
    vector<int>v(n);
    int leftindex=0,rightindex=n-1,sum=-1,temp=0,tempindex=0;
    for(int i=0;i<n;i++){
        cin>>v[i];
        temp+=v[i];
        if(temp<0){
            temp=0;
            tempindex=i+1;
        }
        else if(temp>sum){
            sum=temp;
            leftindex=tempindex;
            rightindex=i;
        }
    }
    if(sum<0) sum=0;
    printf("%d %d %d",sum,v[leftindex],v[rightindex]);
    return 0;
}