P3009 [USACO11JAN]利润Profits

向右看齐 2021-11-26 07:36 246阅读 0赞

## 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。

## 输入输出格式 ##

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 ##

输入样例\#1：  复制

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

输出样例\#1：  复制

## 说明 ##

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。

#include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<cmath>
    #include<queue>
    using namespace std;
    int dp[100001],a[100001],n;
    int main(){
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        for(int i=1;i<=n;i++){
            dp[i]=max(dp[i-1]+a[i],a[i]);
        }
        sort(dp+1,dp+n+1);
        printf("%d",dp[n]);
        return 0;
    }

转载于:https://www.cnblogs.com/xiongchongwen/p/11206062.html