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