Eugeny and Play List

 Eugeny and Play List

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Eugeny loves listening to music. He has n songs in his play list. We know that song number i has the duration of t_i minutes. Eugeny listens to each song, perhaps more than once. He listens to song number i c_i times. Eugeny's play list is organized as follows: first song number 1 plays c₁ times, then song number 2 plays c₂ times, ..., in the end the song number n plays c_n times.

Eugeny took a piece of paper and wrote out m moments of time when he liked a song. Now for each such moment he wants to know the number of the song that played at that moment. The moment x means that Eugeny wants to know which song was playing during the x-th minute of his listening to the play list.

Help Eugeny and calculate the required numbers of songs.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 10⁵). The next n lines contain pairs of integers. The i-th line contains integers c_i, t_i(1 ≤ c_i, t_i ≤ 10⁹) — the description of the play list. It is guaranteed that the play list's total duration doesn't exceed 10⁹ .

The next line contains m positive integers v₁, v₂, ..., v_m, that describe the moments Eugeny has written out. It is guaranteed that there isn't such moment of time v_i, when the music doesn't play any longer. It is guaranteed that v_i < v_i + 1 (i < m).

The moment of time v_i means that Eugeny wants to know which song was playing during the v_i-th munite from the start of listening to the playlist.

Output

Print m integers — the i-th number must equal the number of the song that was playing during the v_i-th minute after Eugeny started listening to the play list.

Sample test(s)

input

1 22 81 16

output

11

input

4 91 22 11 12 21 2 3 4 5 6 7 8 9

output

112234444

题意:有按顺序播放的CD,每张播放C次,每次播放T分钟,给一个时间点,求此时的播放CD序号

思路:因为CD播放的时间宽度达10^9,所以无法数组直接保存下来,但是总的时间宽度也同时只为10^9,而且CD总共只有10^5个,那么只要记载下相应的时间点,也就是最多只有10^5+1个,如果直接查询的话,O(N)的算法,但是有m=10^5组询问数组,TLE,所以二分查询即可,log(n)秒杀,因为是单调有序的上升时刻.

ＡＣ Ｐｒｏｇｒａｍ：

#include <algorithm>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>typedef long long ll;#defineclr(a)memset((a),0,sizeof (a))#definerep(i,a,b)for(int i=(a);i<(int)(b);i++)#defineper(i,a,b)for(int i=((a)-1);i>=(int)(b);i--)#defineinf0x7ffffff#defineeps1e-6using namespace std;int d[100005];int flag;void binary_search(int q,int a,int b){//jump out while a=b-1     if(flag)return;     if(d[a]>q || d[b]<q)return;      if(a==b-1){        cout<<b<<endl;        flag=1;        return;                }     binary_search(q,a,(a+b)/2);//binary_search(q,a,b/2);     binary_search(q,(a+b)/2,b);                              }int main(){  //保持n+1段,在n和n+1段则是属于n的标号的cd   int n,m;  cin>>n>>m;  d[0]=1;  int kg=1;  int c,t;  int sum=0;  for(int i=0;i<n;i++){     cin>>c>>t;     sum+=c*t;     d[kg++]=sum;       }  int q;  while(m--){     cin>>q;             flag=0;      binary_search(q,0,kg-1);       }  //system("pause");  return 0;}