题目描述
Recently Jack becomes much more romantic. He would like to prepare several bunches of flowers.
Each bunch of flowers must have exactly M flowers. As Jack does not want to be boring, he hopes that flowers in the same bunch are all different species. Now there are {N}N species of flowers in the flower shop, and the number of the i-th species of flower is a_ia
i
. Now Jack would like to know how many bunches of flowers he can prepare at most.
\textit{(Flowers are used to propose.)}(Flowers are used to propose.)
输入描述:
The first line contains an integer {T}T (1 \le T \le 101≤T≤10) --- the number of test cases.
In the first line of each test case, there are two integers {N}N, {M}M (1 \le N, M \le 300,0001≤N,M≤300000) --- the number of flowers' species and the number of flowers in a bunch.
In the second line of each test case, there are {N}N integers --- the {i}i-th integer indicates ai the number of {i}i-th species' flowers.
输出描述:
For each test case, output one integer in one line --- the answer of the corresponding test case.
示例1
输入
复制
1 5 3 1 1 1 2 1
输出
复制
2
题意:
一共有n种花,每种花数量为a[i],要用这些花来做成花束,每个花束必须正好有M多花,且都是不同品种,问最多能做成多少束花
题解:
假设能做成x束花,那么就需要花的总量为xm,一共有n种花,如果a[i]>x,也就是这种花可以用在每一束,也就是第i种花最多用x个,如果a[i]<x,那第i种花就要全部用完才可以。
我们用tot来记录在x个花束的情况下,现有的能提供多少花
也就是看当前x的情况下,每一种花所能做的贡献是多少,tot为贡献和
如果tot>xm,即供给大于需求,说明情况成立,最佳答案肯定大于等于x
如果tot<xm,即供给小于需求,说明情况不成立,组价答案肯等小于等于x
这样x我们就可以用二分来确定,条件的判断即tot与xm的关系
代码:
#include<cstdio> #include<iostream> #include<queue> #include<set> #include<algorithm> using namespace std; const int maxn=4e5+8; typedef long long ll; ll a[maxn]; int n,m; bool check(ll x)//x束花 { ll tot=0; for(int i=1;i<=n;i++)tot+=min(a[i],x); if(tot>=x*m)return 1; return 0; } int main() { ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int t; cin>>t; while(t--) { cin>>n>>m; ll sum=0; for(int i=1;i<=n;i++) { cin>>a[i]; sum+=a[i]; } ll l=0,r=sum/m,ans; while(l<=r) { ll mid=(l+r)>>1; if(check(mid)) { l=mid+1; ans=mid; } else r=mid-1; } cout<<ans<<endl; } return 0; }