luogu3009 [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.

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.

贪心即可

注意maxp赋初值要小一点

-1就会wa一个点

#include<bits/stdc++.h>
using namespace std;
template <typename T> void read(T &x){
	x=0;int f=1;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	x*=f;
}

int main(){
	int n,x,maxp=-1e9,sum=0;
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%d",&x);
		if(sum>0) sum+=x;
		else sum=x;
		maxp=max(maxp,sum);
	}
	printf("%d\n",maxp);
 	return 0;
}