洛谷-P3009 [USACO11JAN]Profits S

洛谷-P3009 [USACO11JAN]Profits S

原题链接：https://www.luogu.com.cn/problem/P3009

• 题目描述
• 输入格式
• 输出格式
• 输入输出样例
• 说明/提示
• C++代码

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题目描述

The cows have opened a new business, and Farmer John wants to see how well they are doing. The business has been running for N (1 <= N <= 100,000) days, and every day i the cows recorded their net profit P_i (-1,000 <= P_i <= 1,000).

Farmer John wants to find the largest total profit that the cows have made during any consecutive time period. (Note that a consecutive time period can range in length from one day through N days.) Help him by writing a program to calculate the largest sum of consecutive profits.

奶牛们开始了新的生意，它们的主人约翰想知道它们到底能做得多好。这笔生意已经做了N（1≤N≤100,000）天，每天奶牛们都会记录下这一天的利润Pi（-1,000≤Pi≤1,000）。

约翰想要找到奶牛们在连续的时间期间所获得的最大的总利润。（注：连续时间的周期长度范围从第一天到第N天）。

请你写一个计算最大利润的程序来帮助他。

输入格式

* Line 1: A single integer: N

* Lines 2..N+1: Line i+1 contains a single integer: P_i

输出格式

* Line 1: A single integer representing the value of the maximum sum of profits for any consecutive time period.

输入输出样例

输入 #1

7 
-3 
4 
9 
-2 
-5 
8 
-3 

输出 #1

14 

说明/提示

The maximum sum is obtained by taking the sum from the second through the sixth number (4, 9, -2, -5, 8) => 14.

感谢@smartzzh 提供的翻译。

C++代码

#include <iostream>
using namespace std;

int dp[100005];

int main() {
    int n, ans=0;
    cin >> n;
    int p[n];
    for (int i=0; i<n; ++i)
        cin >> p[i];
    dp[0] = p[0]>0?p[0]:0;
    for (int i=1; i<n; ++i) {
        dp[i] += dp[i-1] + p[i];
        if (dp[i] < 0)
            dp[i] = 0;
        if (ans < dp[i])
            ans = dp[i];
    }
    if (ans == 0) {
        ans = p[0];
        for (int i=1; i<n; ++i)
            if (ans < p[i])
                ans = p[i];
    }
    cout << ans << endl;
    return 0;
}