英文题面：

De Prezer loves movies and series. He has watched the
Troy for like 100 times and also he is a big fan of
Supernatural series.So, he did some researches and found a cursed object which had
n lights on it and initially all of them were turned off.Because of his love to the
Troy, he called that object
Troy.

He looked and saw a note on it in
Khikulish (language of people of Khikuland): "Ma in hame rah umadim, hala migi ... ? ... Mage man De Prezer am k mikhay mano ... ? .... Man se sale ... To boro .... beshur manam miram ... o mishuram".

He doesn't know Khikulish, so just ignored the note and tested the
Troy.

He realized that the light number
i stays turned on for exactly a_i seconds, and then it turns itself off (if it is turned on, in time
t, in time t + a_i - 1 it will be turned on, but on time
t + a_i it won't be) and the next light will be turned on (if
i < n, next light is the light number
i + 1, otherwise it is light with number
1).

For example if n = 2 and we turn on the first light in time
0, it will be turned on in hole interval
[0, a₁) and in hole interval
[a₁, a₁ + a₂) the second light will be turned on and so on.

In time 0 he turns on the light number
1.

De Prezer also loves query.So he gives you
q queries.In each query he will give you integer
t (the time a revengeful ghost attacked him) and you should print the number of the light that is turned on, in time
t.

The first line of input contains two integers
n and q.

The second line contains n space separated integers,
a₁, a₂, ..., a_n .

The next q lines, each line contains a single integer
t .

1 ≤ n, q ≤ 10⁵

1 ≤ a_i ≤ 10⁹

1 ≤ t ≤ 10¹⁸

For each query, print the answer in one line.

        就是说有n盏灯，然后给出每盏灯亮的时间（第一盏从时刻0開始亮直到时刻0+a1-1，第二盏在a1时刻亮起，以此类推），然后题目给出q组询问，询问内容是一个t,要你给出t时刻亮的时哪盏灯。

Sample test(s)：

5 7

1 2 3 4 5

1

2

3

7

14

15

16

2

2

3

4

5

1

2

        这组測试数据还是业界良心的。爆出了可能出错的坑，这一串灯是循环的首尾相接地亮。所以那个t是10^18这么大，要取模。

代码：

#include <iostream>

#include<stdio.h>

using namespace std;

long long int start[100005];//每盏灯開始亮时刻

long long int End[100005];//每盏灯亮完时刻

int n,q,a;

long long int t;

int work(long long int ask)//每次二分查找函数

{

    int p=1;

    int p1=1;

    int p2=n;

    while(ask<start[p]||ask>End[p])

    {

        p=(p1+p2)/2;

        if(ask>=start[p]&&ask<=End[p])

        {

            break;

        }else if(ask>=start[p1]&&ask<=End[p1])

        {

            p=p1;

            break;

        }else if(ask>=start[p2]&&ask<=End[p2])

        {

            p=p2;

            break;

        }else if(ask<start[p]&&ask>End[p1])

        {

            p2=p;

        }else if(ask>End[p]&&ask<start[p2])

        {

            p1=p;

        }

    }

    return p;//p就是那盏灯

}

int main()

{

    scanf("%d%d",&n,&q);

    start[1]=0;

    for(int i=1;i<=n;i++)

    {

        scanf("%d",&a);

        End[i]=start[i]+a-1;

        start[i+1]=start[i]+a;

    }

    while(q--)

    {

        scanf("%I64d",&t);

        printf("%d\n",work(t%start[n+1]));

    }

    return 0;

}