最大连续和

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.

解法

依次扫描。currsum保存上一个数结尾的最大和，maxsum保存全局最大和。那么碰到一个新的数x，有两种行为，一种是以它为头建立一个新的连续序列，一种是把它放在上一个数所在的序列的结尾，看哪个得到的和大一点就用哪个。

注意初始化要比数据范围下界小。

#include <algorithm>
#include <iostream>

using namespace std;

int main(void) {
    int n;
    cin >> n;
    
    int currsum = -2000;
    int maxsum  = -2000;
    
    for (int i = 0; i < n; ++i) {
        int x;
        cin >> x;
        
        currsum = max(x, x+currsum);
        maxsum  = max(maxsum, currsum);
    }
    
    cout << maxsum;
}