洛谷-P3009 [USACO11JAN]Profits S
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.
奶牛们开始了新的生意，它们的主人约翰想知道它们到底能做得多好。这笔生意已经做了N（1≤N≤100,000）天，每天奶牛们都会记录下这一天的利润Pi（-1,000≤Pi≤1,000）。
约翰想要找到奶牛们在连续的时间期间所获得的最大的总利润。（注：连续时间的周期长度范围从第一天到第N天）。
请你写一个计算最大利润的程序来帮助他。
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 #1
7
-3
4
9
-2
-5
8
-3
输出 #1
14
说明/提示
The maximum sum is obtained by taking the sum from the second through the sixth number (4, 9, -2, -5, 8) => 14.
感谢@smartzzh 提供的翻译。
C++代码
#include <iostream>
using namespace std;
int dp[100005];
int main() {
int n, ans=0;
cin >> n;
int p[n];
for (int i=0; i<n; ++i)
cin >> p[i];
dp[0] = p[0]>0?p[0]:0;
for (int i=1; i<n; ++i) {
dp[i] += dp[i-1] + p[i];
if (dp[i] < 0)
dp[i] = 0;
if (ans < dp[i])
ans = dp[i];
}
if (ans == 0) {
ans = p[0];
for (int i=1; i<n; ++i)
if (ans < p[i])
ans = p[i];
}
cout << ans << endl;
return 0;
}