Recently, Bob has just learnt a naive sorting algorithm: merge sort. Now, Bob receives a task from Alice.
Alice will give Bob N sorted sequences, and the i-th sequence includes ai elements. Bob need to merge all of these sequences. He can write a program, which can merge no more than k sequences in one time. The cost of a merging operation is the sum of the length of these sequences. Unfortunately, Alice allows this program to use no more than T cost. So Bob wants to know the smallest k to make the program complete in time.
Input
The first line of input contains an integer t0, the number of test cases. t0 test cases follow.
For each test case, the first line consists two integers N (2≤N≤100000) and T (∑Ni=1ai<T<231).
In the next line there are N integers a1,a2,a3,...,aN(∀i,0≤ai≤1000).
For each test case, the first line consists two integers N (2≤N≤100000) and T (∑Ni=1ai<T<231).
In the next line there are N integers a1,a2,a3,...,aN(∀i,0≤ai≤1000).
Output
For each test cases, output the smallest k.
Sample Input
1
5 25
1 2 3 4 5
Sample Output
3
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstdlib> 5 #include <cstring> 6 #include <string> 7 #include <deque> 8 #include <queue> 9 using namespace std; 10 #define ll long long 11 #define N 100009 12 #define gep(i,a,b) for(int i=a;i<=b;i++) 13 #define gepp(i,a,b) for(int i=a;i>=b;i--) 14 #define gep1(i,a,b) for(ll i=a;i<=b;i++) 15 #define gepp1(i,a,b) for(ll i=a;i>=b;i--) 16 #define mem(a,b) memset(a,b,sizeof(a)) 17 int n,m,t; 18 int sum[N],a[N]; 19 priority_queue<int,vector<int>,greater<int> >que; 20 bool check(int k){ 21 while(!que.empty()) que.pop(); 22 int x=(n-1)%(k-1);//共需要归并n-1个数，每次要归并k-1个数 23 int add=0; 24 if(x){//为了不影响单调性 25 x++;//每一次都是减去X个，又加一个。 26 add+=sum[x]; 27 que.push(add);//先去掉x个，后面正常 28 } 29 gep(i,x+1,n) que.push(a[i]); 30 int y=(n-1)/(k-1);//分步骤 31 gep(i,0,y-1){ 32 int tmp=k; 33 int tt=0; 34 while(tmp--){//每次k个 35 int v=que.top(); 36 que.pop(); 37 tt+=v; 38 } 39 que.push(tt);//再压入 40 add+=tt; 41 } 42 return add<=m;//不超过m才行 43 } 44 int main() 45 { 46 scanf("%d",&t); 47 while(t--){ 48 scanf("%d%d",&n,&m); 49 mem(a,0); 50 mem(sum,0); 51 gep(i,1,n) { 52 scanf("%d",&a[i]); 53 } 54 sort(a+1,a+1+n);//要先排序 55 gep(i,1,n){ 56 sum[i]=sum[i-1]+a[i]; 57 } 58 int l=2,r=n;//至少2个 59 while(l<=r){//r可能取到 60 int mid=(r+l)>>1; 61 if(check(mid)) r=mid-1;//r=mid是错的，会死循环 62 else l=mid+1; 63 } 64 printf("%d ",r+1); 65 } 66 return 0; 67 }