Maximum|Maximum Subsequence Sum MaximumSubsequenceSum

https://pintia.cn/problem-sets/900290821590183936/problems/900291257604861953
复杂度2 Maximum Subsequence Sum
【Maximum|Maximum Subsequence Sum】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

import java.io.BufferedReader; import java.io.IOException; import java.io.InputStream; import java.io.InputStreamReader; import java.util.StringTokenizer; class Reader { static BufferedReader reader; static StringTokenizer tokenizer; static void init(InputStream input) { reader = new BufferedReader( new InputStreamReader(input) ); tokenizer = new StringTokenizer(""); }static String next() throws IOException { while ( ! tokenizer.hasMoreTokens() ) {tokenizer = new StringTokenizer( reader.readLine() ); } return tokenizer.nextToken(); }static int nextInt() throws IOException { return Integer.parseInt( next() ); }static double nextDouble() throws IOException { return Double.parseDouble( next() ); } }public class Main {public static void main(String[] args) throws IOException { Reader.init( System.in ); int k=Reader.nextInt(); int csum=0; int sum=-1; int x=0; int t=0; int y=-1; int s[]=new int[10001]; for(int i=0; isum){ sum=csum; x=t; y=i; } if(csum<0){ csum=0; t=i+1; } } if(y==-1){ System.out.printf("0 %d %d\n",s[0],s[k-1]); }else { System.out.printf("%d %d %d\n",sum,s[x],s[y]); }} }

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)); br.readLine(); String s = br.readLine().trim(); String[] nums = s.split(" "); int thisSum = 0; int maxSum = 0; int i = 0; //smallest i int j = 0; //smallest j /* * How to find the smallest i ? */ boolean isNegative = true; boolean isFirst = true; int thisi = 0; for(int start = 0; start < nums.length; start++) { int temp = Integer.parseInt(nums[start]); if(isNegative && temp >= 0) isNegative = false; thisSum += temp; if(isFirst && thisSum >= 0) { thisi = temp; isFirst = false; } if(thisSum > maxSum) { i = thisi; maxSum = thisSum; j = temp; } else if(thisSum < 0) { thisSum = 0; isFirst = true; } } if(isNegative) System.out.println(0 + " " + nums[0] + " " + nums[nums.length - 1]); else System.out.println(maxSum + " " + i + " " + j); } }

#include using namespace std; int main() { int k,csum=0,sum=-1,x=0,t=0,y=-1; //t 表示成为下一个最大子序列的第一个节点 int s[10001]; scanf("%d",&k); for(int i=0; isum){ sum=csum; x=t; y=i; } if(csum<0){ csum=0; t=i+1; } } if(y==-1)printf("0 %d %d\n",s[0],s[k-1]); else printf("%d %d %d\n",sum,s[x],s[y]); }