标签：integer pop watch code i++ printf using return n+1
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\n",work(t%start[n+1]));
}
return 0;
}
Codeforces Hello2015第一题Cursed Query
标签：integer pop watch code i++ printf using return n+1
原文地址：http://www.cnblogs.com/lytwajue/p/7222928.html
