PAT甲级1007

1007 Maximum Subsequence Sum (25point(s))

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

题目大意：给一串数字，求最大连续子序列和。
解题思路：假设这个数组为A[]，那么我们用dp[i]表示以A[i]结尾连续序列的最大和。有以下两种情况：①这个最大和的连续序列只有一个元素，即A[i]。②这个最大和的连续序列有多个元素，即从前面某处A[p]开始(p ，边界为dp[0]=A[0]。

Java代码：

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String s = br.readLine();
		int n = Integer.parseInt(s);
		s = br.readLine();
		String str[] = s.split(" ");
		int arr[] = new int[n];
		for(int i=0; imaxn)
			{
				maxn = sum;
				left = temp;
				right = i;
			}
		}
		if(maxn<0) maxn=0;
		System.out.println(maxn+" "+arr[left]+" "+arr[right]);
	}
}

注意：在使用Java解题时，测试用例数据量可能会很大时，读入操作如果使用Scanner类，会出现超时的情况，因为Scanner的读入操作效率极低，在读入数据时可能就已经超时，这里推荐使用BufferedReader类，读入速度将远高于Scanner，当然也有一定的弊端，BufferReader不能直接读入数字，这里可以采用读入一行数据后再对字符串分割转换成整数数组，相比C++会较为繁琐，这也是为什么算法题目推荐使用C++而不用Java的原因。

C++代码：

#include 
using namespace std;
int main()
{
    int n;
    scanf("%d",&n);
    vector A(n);
    for(int i=0; imaxn)
        {
            maxn = sum;
            left = temp;
            right = i;
        }
    }
    if(maxn<0) maxn=0;
    printf("%d %d %d", maxn, A[left], A[right]);
}