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;
}