首页 > 试题广场 >

Maximum Subsequence Sum (25)

[编程题]Maximum Subsequence Sum (25)

热度指数：5510 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 32M，其他语言64M
算法知识视频讲解

Given a sequence of K integers { N₁ , N₂ , ..., 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.

输入描述:

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.

输出描述:

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.

示例1

输入

10<br/>-10 1 2 3 4 -5 -23 3 7 -21

输出

10 1 4

算法知识视频讲解

Java

HackerLzh

核心是一道经典的DP题

import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int K = Integer.parseInt(br.readLine());
        int[] nums = new int[K];
        String[] input = br.readLine().split("\\s+");
        int nonNegative = 0;
        for (int i = 0; i < K; ++i) {
            int tmp = Integer.parseInt(input[i]);
            nums[i] = tmp;
            if (tmp >= 0) {
                nonNegative++;
            }
        }
        input = null;
        if (nonNegative == 0) {
            System.out.format("%d %d %d\n", 0, nums[0], nums[K - 1]);
            return;
        }

        int maxSum = nums[0];
        int maxSumIndex = 0;
        int maxSumLen = 1;
        int preMaxSum = nums[0];
        int preMaxSumLen = 1;
        for (int i = 1; i < K; ++i) {
            if (preMaxSum >= 0) {
                preMaxSum += nums[i];
                preMaxSumLen++;
            } else {
                preMaxSum = nums[i];
                preMaxSumLen = 1;
            }
            if (nums[i] >= 0 && preMaxSum > maxSum) {
                maxSum = preMaxSum;
                maxSumIndex = i;
                maxSumLen = preMaxSumLen;
            }
        }
        System.out.format("%d %d %d\n", maxSum, nums[maxSumIndex - maxSumLen + 1], nums[maxSumIndex]);

    }
}

发表于 2025-02-11 15:39:18 回复(0)

Java

DaaAMillian

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

public class Main {
private static int N;
private static int[] sum;
private static int[] start;
private static int[] a;

public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

String str;
String[] strs;

while((str=br.readLine())!=null) {

//read data
N = Integer.valueOf(str);

sum = new int[N];
start = new int[N];
a = new int[N];

strs = br.readLine().split(" ");
for(int i=0; i<N; i++) {
a[i] = Integer.valueOf(strs[i]);
}

//init
sum[0] = a[0];
start[0] = 0;

//iterative
for (int i = 1; i < N; i++) {
if (sum[i - 1] > 0) {
sum[i] = sum[i - 1] + a[i];
start[i] = start[i - 1];
} else {
sum[i] = a[i];
start[i] = i;
}
}

//reconstruction solution
int max = Integer.MIN_VALUE;
int index = 0;
for (int i = 0; i < N; i++) {
if (max < sum[i]) {
max = sum[i];
index = i;
}
}

if (max < 0) {
System.out.println(0 + " " + a[0] + " " + a[N - 1]);
} else {
System.out.println(max + " " + a[start[index]] + " " + a[index]);
}
}
}
}

为什么牛客有两个case没过，它也没给全我没过的case，不知道哪里出现问题，PAT都过了

发表于 2018-09-26 17:26:27 回复(0)