【 USACO11JAN】 利润 【洛谷P3009】

P3009 [USACO11JAN]利润Profits

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.

牛们开了家新公司，这家公司已经运作了N天，财务报表显示第i天获得的利润为Pi ， 有些天的利润可能是个负数。约翰想给奶牛公司写个新闻报道，以吹嘘她们的业绩。于是他 想知道，这家公司在哪一段连续的日子里，利润总和是最大的。

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：

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

输出样例#1：

14 

说明

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

这道题很裸，但是思想和方法非常重要，尽管比较简单。

不给详细解释，请大家仔细研读代码，如有不明，私信或评论或Q我（568251782）均可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>

const int MAXN  = 100000 + 10;
const int INF = 99999999;

int n,ans;
int num[MAXN];


int main()
{
	scanf("%d", &n);
	ans = -1*INF;
	for(int i= 1;i <= n;i++)
	{
		scanf("%d", &num[i]);
		if(num[i-1] > 0)
		{
			num[i] += num[i-1];
		}
		ans = std::max(ans, num[i]);
	}
	printf("%d", ans);
	return 0;
}