題目大意：將一個個數為n的序列分割成m份，要求這m份中的每份中值（該份中的元素和）最大值最小， 輸出切割方式，有多種情況輸出使得越前面越小的情況。

解題思路：二分法求解f(x), f(x) < 0 說明不能滿足， f(x) >= 0說明可以滿足，f(x) 就是當前最大值為x的情況最少需要劃分多少份-要求份數（如果f(x ) >= 0 說明符合要求而且還過於滿足，即x還可以更小）。

注意用long long .

#include <stdio.h>
#include <string.h>
int max(const int &a, const int &b) { return a > b ? a : b;}

const int N = 1005;
int n, T;
long long num[N], sum[N], rec[N];

bool judge(int Max) {
    int cnt = 0, first = 0, end = 1;
    while (end <= n) {
	if (sum[end] - sum[first] > Max) {
	    cnt++;
	    first = end - 1;
	}
	end++;
    }
    return cnt <= T - 1;
}

int main () {
    long long  cas, lift, right, cur;
    scanf("%lld", &cas);
    while (cas--) {
	// Init;
	memset(num, 0, sizeof(num));
	memset(sum, 0, sizeof(sum));
	memset(rec, 0, sizeof(rec));
	lift = right = 0;

	// Read.
	scanf("%d%d", &n, &T);
	for (int i = 1; i <= n; i++) {
	    scanf("%lld", &num[i]);
	    sum[i] = sum[i - 1] + num[i];
	    lift = max(lift, num[i]);
	}
	right = sum[n];

	// Count;
	while (lift != right) {
	    cur = (right + lift) / 2;
	    if (judge(cur)) {
		right = cur;
	    }
	    else
		lift = cur + 1;
	}

	// Draw;
	long long now = 0, cnt = 0;
	for (int i = n; i > 0; i--) {
	    if (now + num[i] > lift || i < T - cnt) {
		rec[i] = 1;
		cnt++;
		now = num[i];
	    }
	    else
		now += num[i];
	}

	// Printf;
	for (int i = 1; i < n; i++) {
	    printf("%lld ", num[i]);
	    if (rec[i])	    printf("/ ");
	}
	printf("%lld\n", num[n]);
    }
    return 0;
}