传送门

描述

YaoYao has a company and he wants to employ m people recently. Since his company is so famous, there are n people coming for the interview. However, YaoYao is so busy that he has no time to interview them by himself. So he decides to select exact m interviewers for this task.

YaoYao decides to make the interview as follows. First he queues the interviewees according to their coming order. Then he cuts the queue into m segments. The length of each segment is n/m, which means he ignores the rest interviewees (poor guys because they comes late). Then, each segment is assigned to an interviewer and the interviewer chooses the best one from them as the employee.

YaoYao’s idea seems to be wonderful, but he meets another problem. He values the ability of the ith arrived interviewee as a number from 0 to 1000. Of course, the better one is, the higher ability value one has. He wants his employees good enough, so the sum of the ability values of his employees must exceed his target k (exceed means strictly large than). On the other hand, he wants to employ as less people as possible because of the high salary nowadays. Could you help him to find the smallest m?

输入

The input consists of multiple cases.

In the first line of each case, there are two numbers n and k, indicating the number of the original people and the sum of the ability values of employees YaoYao wants to hire (n≤200000, k≤1000000000). In the second line, there are n numbers v1, v2, …, vn (each number is between 0 and 1000), indicating the ability value of each arrived interviewee respectively.

The input ends up with two negative numbers, which should not be processed as a case.

输出

For each test case, print only one number indicating the smallest m you can find. If you can’t find any, output -1 instead

样例

• 输入
  11 300
  7 100 7 101 100 100 9 100 100 110 110
  -1 -1
• 输出
  3

思路

• 题意：给出n个数和一个k，要求把这n个数分成m段，多出来的不要，使得这m段中每一段的最大值加起来大于k，同时要求m尽可能小，求m。
• 所有数的和小于k则无解，整个序列的最大值maxval大于k则m=1，m从max(2,k/maxval)开始递增，对于每个m，找出每一段的最大值和，直到其大于k。
• 一维ST表存最值。

Code

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#include
         
           #include
          
            #include
           
             #include
            
              #define INIT(a,b) memset(a,b,sizeof(a)) #define LL long long int using namespace std; const int MAX=0x7fffffff; const int MIN=-0x7fffffff; const int INF=0x3f3f3f3f; const int Mod=1e9+7; const int maxn=200007; int n,k,num[maxn],st[maxn][30]; void Init(){
            for(int j=1;(1<
             
              <=n;j++)         for(int i=1;i+(1<
              
               <=n;i++) st[i][j]=max(st[i][j-1],st[i+(1<<(j-1))][j-1]); } int Query(int l,int r){ int k=log(r-l+1.0)/log(2.0); return max(st[l][k],st[r-(1<
               
                =k){ printf("1\n"); continue; } Init(); int m=max(2,k/maxval);//m最小值 while(m<=n){ int ans=0,x=n/m; for(int i=1;i<=m;i++) ans+=Query((i-1)*x+1,i*x); if(ans>k){ printf("%d\n",m); break; } m++; } } return 0; }