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[USACO11JAN]利润Profits

时间:2016-12-04 19:46:57      阅读:274      评论:0      收藏:0      [点我收藏+]

标签:space   running   输出   一段   represent   from   using   names   new   

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:

 1 #include<cstdio>
 2 #include<iostream>
 3 using namespace std;
 4 int n,a,b,c,l=-3000,f[2][2];
 5 int main(){
 6     scanf("%d",&n);
 7     for(int i=1;i<=n;i++){
 8         scanf("%d",&a);
 9         b=i%2;c=(b+1)%2;l=max(l,a);
10         f[b][0]=max(f[c][0],f[c][1]);
11         f[b][1]=max(f[c][1]+a,a);
12     }
13     if(l>=0) printf("%d\n",max(f[n%2][0],f[n%2][1]));
14     else printf("%d\n",l);
15     return 0;
16 }

還記得某種數列問題嗎?

[USACO11JAN]利润Profits

标签:space   running   输出   一段   represent   from   using   names   new   

原文地址:http://www.cnblogs.com/J-william/p/6131251.html

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