Google Code Jam 2015 Round 1A Haircut 二分

題意：給你每個理發師的理發時間，問你排在隊列中的第N個位置，問你應該被哪個理發師剪發。

解題思路：二分時間，看這個時間到第幾個人理發了，然后找到臨界值，看這個值的時候有那些理發師接待了新旅客的，即可找到那個理發師。

解題代碼：

// File Name: b.sample.cpp
// Author: darkdream
// Created Time: 2015年04月18日 星期六 09時05分30秒

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<ctime>
#define LL long long

using namespace std;
LL a[1005];
LL b , n ; 
LL count(LL t){
    LL sum = 0 ; 
    for(int i = 1;i <= b;i ++)
    {
      sum += (t/a[i]) + 1 ;    
    }
    return sum;
}
LL fen(LL l , LL  r){
    LL m = (l + r)/2;
    while(l<= r){
        //printf("***\n");
         m = (l + r)/2;             
         LL t = count(m);
         if(t >= n){
            r = m - 1; 
         }else l = m + 1; 
    }
    return l ; 
}
int main(){
//    freopen("B-large.in","r",stdin);
//    freopen("output","w",stdout);
    LL T;
    scanf("%lld",&T);
    for(int ca = 1; ca <= T ;ca ++){
        scanf("%lld %lld",&b,&n);
        for(LL i = 1;i <= b;i ++)
            scanf("%lld",&a[i]);
        printf("Case #%d:",ca);
        if(n <= b  )
        {
            printf(" %lld\n",n);
            continue;
        }
        LL t = fen(1,1e15);
        LL num = count(t-1);
        for(int i = 1;i <= b;i ++){
            if(t% a[i] == 0){
                num ++;
            }
            if(num == n)
            {
                printf(" %d\n",i);
                break;
            }
        }
        
    }

return 0;
}

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