原題：

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 ) that may have different number of pages () and you want to make one copy of each of them. Your task is to divide these books among k scribes, . 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  such that i-th scriber gets a sequence of books with numbers between b_i-1+1 and b_i. 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, . At the second line, there are integers  separated by spaces. All these values are positive and less than 10000000.

For each case, print exactly one line. The line must contain the input succession  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.

2
9 3
100 200 300 400 500 600 700 800 900
5 4
100 100 100 100 100

100 200 300 400 500 / 600 700 / 800 900
100 / 100 / 100 / 100 100

題目大意：

要抄N本書，編號為1,2,3...N, 每本書有1<=x<=10000000頁， 把這些書分配給K個抄寫員，要求分配給某個抄寫員的那些書的編號必須是連續的。每個抄寫員的速度是相同的，求所有書抄完所用的最少時間的分配方案。

分析與總結：

所化的總時間取決於所有抄寫員中任務最多的那個，是經典的最大值最小化問題。LRJ《演算法入門經典》P151頁有介紹：

/**
*  最大值最小化問題： 二分法。  劉汝佳那本白書上也說的很詳細。
*     先把最大值的可能區間計算出來，也就是[0, maxAns] maxAns 就是所有頁數總和。
*  然後再在答案區間裡先把“最大值最小化”。
*     得到最小化後的最大值以後，就可以根據這個最大值來對最終答案進行計算。
*  如何分割槽，也就是計算最後答案ans：  這裡就要用到貪心思想，因為題目要求
*  如果有多種可能情況，讓區間集合所含的元素數從小到大排列。這樣就可以貪心地
*  從所有pages的最後一個元素往前遍歷來進行劃分區間，只要其總和不大於之前所求的最大值，
*  就歸到那個區間。
*     在把所有區間分好後，還有個情況就是，這個時候可能需要3個斜槓來分4個區間，
*  然而只用了2個（比如樣例2），這樣還有一個斜槓沒用到，就再用貪心，把pages從第一個開始往後掃，
*  如果當前pages不用劃斜槓，而剩餘斜槓數大於0，則在這裡添個斜槓。
*/

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#define INF 0x7fffffff
#define MAXS 501
#define LL long long
using namespace std;
int m, k;
LL maxAns;
int pages[MAXS], ans[MAXS];

bool judge(LL x) {
    int sum = 0, t = k;
    for(int i = 0; i < m; i ++) {
        sum += pages[i];
        if(sum > x) {
            i --;
            sum = 0;
            t --;
        }
        if(!t) {
            if(i != m - 1) return false;
            else return true;
        }
    }
    return true;
}

void solve() {
    memset(ans, 0, sizeof(ans));
    LL l, r, cur;
    l = 0; r = maxAns;
    while(l < r) {
        cur = (l + r) / 2;
        if(judge(cur))   r = cur;
        else             l = cur + 1;
    }
    int sum = 0;
    for(int i = m - 1; i >= 0; i --) {
        sum += pages[i];
        if(sum > r) {
            sum = 0;
            ans[++ i] = 1;
            k --;
        }
    }
    int i = 1;
    while(k > 1) {
        for(; i < m; i ++ ) {
            if(!ans[i]) {
                ans[i] = 1;
                k --;
                break;
            }
        }
    }
    printf("%d", pages[0]);
    for(int i = 1; i < m; i ++) {
        if(ans[i]) printf(" /");
        printf(" %d", pages[i]);
    }
    printf("\n");
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t --) {
        maxAns = 0;
        scanf("%d%d", &m, &k);
        for(int i = 0; i < m; i ++) {
            scanf("%d", &pages[i]);
            maxAns += pages[i];
        }
        solve();
    }
    return 0;
}