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.

簡單動規。(因為方程沒法應付全負的情況,加了個l特判)

代碼實現:

 #include<cstdio>
#include<iostream>
using namespace std;
int n,a,b,c,l=-,f[][];
int main(){
scanf("%d",&n);
for(int i=;i<=n;i++){
scanf("%d",&a);
b=i%;c=(b+)%;l=max(l,a);
f[b][]=max(f[c][],f[c][]);
f[b][]=max(f[c][]+a,a);
}
if(l>=) printf("%d\n",max(f[n%][],f[n%][]));
else printf("%d\n",l);
return ;
}

還記得某種數列問題嗎?

05-08 08:45