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Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had
to be re-written by hand by so called scribers. The scriber had been given a book and after several
months he finished its copy. One of the most famous scribers lived in the 15th century and his name
was Xaverius Endricus Remius Ontius Xendrianus (Xerox). Anyway, the work was very annoying and
boring. And the only way to speed it up was to hire more scribers.
Once upon a time, there was a theater ensemble that wanted to play famous Antique Tragedies. The
scripts of these plays were divided into many books and actors needed more copies of them, of course.
So they hired many scribers to make copies of these books. Imagine you have m books (numbered
1, 2, . . . , m) that may have different number of pages (p1, p2, . . . , pm) and you want to make one copy of
each of them. Your task is to divide these books among k scribes, k ≤ m. Each book can be assigned
to a single scriber only, and every scriber must get a continuous sequence of books. That means, there
exists an increasing succession of numbers 0 = b0 < b1 < b2, . . . < bk−1 ≤ bk = m such that i-th scriber
gets a sequence of books with numbers between bi−1 + 1 and bi
. The time needed to make a copy of
all the books is determined by the scriber who was assigned the most work. Therefore, our goal is to
minimize the maximum number of pages assigned to a single scriber. Your task is to find the optimal
assignment.
Input
The input consists of N cases. The first line of the input contains only positive integer N. Then follow
the cases. Each case consists of exactly two lines. At the first line, there are two integers m and k,
1 ≤ k ≤ m ≤ 500. At the second line, there are integers p1, p2, . . . , pm separated by spaces. All these
values are positive and less than 10000000.
Output
For each case, print exactly one line. The line must contain the input succession p1, p2, . . . pm divided
into exactly k parts such that the maximum sum of a single part should be as small as possible. Use
the slash character (‘/’) to separate the parts. There must be exactly one space character between any
two successive numbers and between the number and the slash.
If there is more than one solution, print the one that minimizes the work assigned to the first scriber,
then to the second scriber etc. But each scriber must be assigned at least one book.
Sample Input
2
9 3
100 200 300 400 500 600 700 800 900
5 4
100 100 100 100 100
Sample Output
100 200 300 400 500 / 600 700 / 800 900
100 / 100 / 100 / 100 100

解题思路：
　　本题的优化目标是使最大连续子序列的和最小，并且在最大子序列和相同的情况下s1、s2...尽量小。那么我们可以从右边开始，尽量向左划分，当目前剩余书本数等于剩余的人数时，剩余每本书的分配策略只能是每人一本。

代码如下：

#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>
#include <algorithm>
using namespace std;
#define maxm 500+5
typedef long long LL;
int m,k;
int p[maxm];
vector<int> s;
int ans[maxm];
LL M;

bool judge(LL x){
    
    s.clear();
    bool flag=true;
    int cnt=k;
    for(int i = m;i>0;){
        LL sum = 0;
        while(i > 0 && sum + p[i] <= x){
            if(i + 1 == cnt) break;
            sum += p[i--];
        }
        s.push_back(i);
        cnt--;
        if(s.size()>k){
            flag = false;
            break;
        }
    }
    if(flag){
        int j=0;
        for(int i = s.size() - 1;i >= 0;i--){
            ans[j++] = s[i];
        }
        return true;
    }
    else return false;
}
int main(int argc, const char * argv[]) {
    freopen("/Users/hujiacheng/Desktop/input.txt", "r", stdin);
    int N;
    scanf("%d",&N);
    while(N--){
        M=0;
        memset(ans, 0, sizeof ans);
        scanf("%d%d",&m,&k);
        for(int i=1;i<=m;i++){
            scanf("%d",&p[i]);
            M += p[i];
        }
        LL l=1,r=M;
        LL mid=(l+r)/2;
        while(l<r){
            if(judge(mid)){
                r=mid;
                mid=(l+r)/2;
                
            }
            else {
                l=mid+1;
                mid=(l+r)/2;
            }
        }
        int j=1;
        for(int i=1;i<=m;i++){
            if(i!=1) cout<<" ";
            cout<<p[i];
            if(i==ans[j]){
                cout<<" /";
                j++;
            }
        }
        cout<<endl;
    }
    return 0;
}