Codeforces Beta Round #6 (Div. 2 Only) C. Alice, Bob and Chocolate 水题

C. Alice, Bob and Chocolate

题目连接：

http://codeforces.com/contest/6/problem/C

Description

Alice and Bob like games. And now they are ready to start a new game. They have placed n chocolate bars in a line. Alice starts to eat chocolate bars one by one from left to right, and Bob — from right to left. For each chocololate bar the time, needed for the player to consume it, is known (Alice and Bob eat them with equal speed). When the player consumes a chocolate bar, he immediately starts with another. It is not allowed to eat two chocolate bars at the same time, to leave the bar unfinished and to make pauses. If both players start to eat the same bar simultaneously, Bob leaves it to Alice as a true gentleman.

How many bars each of the players will consume?

Input

The first line contains one integer n (1 ≤ n ≤ 105) — the amount of bars on the table. The second line contains a sequence t1, t2, ..., tn (1 ≤ ti ≤ 1000), where ti is the time (in seconds) needed to consume the i-th bar (in the order from left to right).

Output

Print two numbers a and b, where a is the amount of bars consumed by Alice, and b is the amount of bars consumed by Bob.

Sample Input

5

2 9 8 2 7

Sample Output

2 3

Hint

题意

有n个物品，每个物品吃掉的时间是a[i]

一个人从左边开始吃，一个人从右边开始吃

如果两个人同时吃到了一个东西，算左边的。

最后问你左边吃了多少个，右边吃了多少个

题解：

直接暴力就好了……

一个记录左边吃的时间，一个记录右边吃的时间，不停去扫就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
long long a[maxn];
int main()
{
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%lld",&a[i]);
    long long t1=0,t2=0;
    int num1=0,num2=0;
    int l=1,r=n;
    while(l<=r)
    {
        if(t1<=t2)t1+=a[l++],num1++;
        else t2+=a[r--],num2++;
    }
    cout<<num1<<" "<<num2<<endl;
}