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复制
7
-3
4
9
-2
-5
8
-3
输出 #1复制
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.
AC Code:
#include<bits/stdc++.h> using namespace std; #define N 100001 int dp[N],a[N]; int n,ans=-0x7fffffff; int main() { scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } for(int i=1;i<=n;i++) { /*对于每个dp[i], 可以从前面的那段加上a[i]得到(dp[i-1]与a[i]是连续的) 也可以自己新起一个子段*/ dp[i]=max(a[i],dp[i-1]+a[i]);//dp[i]表示以a[i]结尾的最大子段和 ans=max(ans,dp[i]);//把每个子段扫描一遍,找到最大子段和 } printf("%d\n",ans); return 0; }