1007 Maximum Subsequence Sum
Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj } where 1. 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 (≤). 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
这破题做的。JAVA O(N)算法都超时。mlgb
思路就是贪心。
import java.util.Scanner; public class Main{ static int Nums[]; static int N; static int Max = 0; static int start; static int end; public static void main(String[] args) { Scanner sc = new Scanner(System.in); N = sc.nextInt(); Nums = new int[N]; int sum = 0;//当前总数 int sign = 0; for(int i=0,j=0;i<N;i++) { Nums[i] = sc.nextInt(); sum+=Nums[i]; if(sum>Max) { Max=sum; start=Nums[j]; end=Nums[i]; } else if(sum<0) { j=i+1; sum=0; sign++; } else if(sum==0&&Max==0) { start=0; end=0; } } if(sign==N) { start=Nums[0]; end=Nums[N-1]; } sc.close(); System.out.println(Max+" "+start+" "+end); } }
用C++改写提交即可
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