2020-10-03

Flowers

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; }