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 ai seconds, and then it turns itself off (if it is turned on, in time t, in time t + ai - 1 it will be turned on, but on time t + ai 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, ..., an .
The next q lines, each line contains a single integer t .
1 ≤ n, q ≤ 105
1 ≤ ai ≤ 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。
我的做法是每次询问就来一次二分查找,log2n的复杂度再盛上10^5次询问能够接受。
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 ",work(t%start[n+1])); } return 0; }