题干：

链接：https://ac.nowcoder.com/acm/contest/886/D
来源：牛客网
 

After the struggle of graduating from college, TangTang is about to move from a student apartment to his new home.

TangTang has n items to move, the i-th of which is of volume viv_ivi​. He can pack all these items into at most K boxes of the same volume.
 

TangTang is so clever that he uses the following strategies for packing items:

 

- Each time, he would put items into a box by the next strategy, and then he would try to fill another box.

- For each box, he would put an unpacked item of the largest suitable volume into the box repeatedly until there is no such item that can be fitted in the box.

Now, the question is what is the minimum volume of these boxes required to pack all items.

输入描述:

There are multiple test cases. The first line contains an integer T (1≤T≤201 \leq T \leq 201≤T≤20), indicating the number of test cases. Test cases are given in the following.For each test case, the first line contains two integers n, K (1≤n,K≤10001 \leq n, K \leq 10001≤n,K≤1000), representing the number of items and the number of boxes respectively.The second line contains n integers v1v_1v1​, v2v_2v2​, …\ldots…, vnv_nvn​ (1≤v1,v2,…,vn≤10001 \leq v_1, v_2, \ldots, v_n \leq 10001≤v1​,v2​,…,vn​≤1000), where the i-th integer viv_ivi​ represents the volume of the i-th item.

输出描述:

For each test case, output "Case #x: y" in one line (without quotes), where x indicates the case number starting from 1, and y denotes the answer to this test case.

示例1

输入

复制

1
5 3
1 2 3 4 5

输出

复制

Case #1: 5

题目大意：

你有n件行李，有k个箱子体积相同的箱子，遵循下面的规则将行李放进箱子里面，每次都取当前最大的可以放进箱子的行李放进箱子，如果该箱子放不进任何行李那么就换一个新的箱子再按照这一条规则进行放行李，请问箱子最小的体积是多少，可以放进所有行李。

解题报告：

打眼一看是二分，但是其实并没有单调性，大概这么理解：假设是这些行李，你如果箱子的体积减小一点，或许就能拿出来一个大行李，放进去更多的小行李。这样导致后面的箱子需要放的行李的序列和 不减小箱子体积的行李序列是完全不一样的了，所以答案肯定也会不同，所以不能二分。

这题的正解是我们可以确定一个下界和一个上界，并且这个界的范围不会很大，所以可以直接在这个上下界内暴力就好。

当然由于不具有二分性质的样例不好构造，我们可以近似认为这是一个具有二分性质的题目，只不过加点随机化的思想。注意这题如果先二分完了在最后进行随机数处理的话也过不去，需要在二分的过程中扩大搜索范围。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;int a[MAX];
int n,k;
bool ok(int x) {multiset<int> ms;int cnt = 0,res = x;for(int i = 1; i<=n; i++) ms.insert(a[i]);while(1) {if(ms.empty() || cnt == k) break;auto it = ms.upper_bound(res);if(it == ms.begin()) {res = x;cnt++;} else {--it;res -= *it;ms.erase(it);}}if(ms.empty()) return 1;else return 0;
}
int main() {int t,iCase=0;cin>>t;while(t--) {scanf("%d%d",&n,&k);for(int i=1; i<=n; i++) scanf("%d",a+i);int l = 1,r=1e6,ans=1e6,mid;while(l+20<=r) {mid=(l+r)>>1;int flag = 0;for(int i = mid; i<=mid+10; i++) {if(ok(i)) {flag = 1,r = i-1,ans=i;break;}}if(flag == 0) l = mid+1;}for(int i = max(0,ans-20); i<=ans; i++) {if(ok(i)) {printf("Case #%d: %d\n",++iCase,i);break;}}}return 0;
}