Uva714-copying books (minimized maximum value)

Question: uva714-copying books (minimum maximum value)

N books are given. The value of each book represents the number of pages in the book. Given M scribers, each scriber must copy at least one book or several successive books. The workload of each scriber is equal to the sum of the pages of the book he wants to copy. Ask how to minimize the maximum workload of scribers. The answer is required to output the preceding and smaller division if there are multiple answers.

Solution: minimize the maximum value.

The possible maximum value of a binary test. If the maximum value can be divided, it means that the maximum value may be this value or smaller. If not, the maximum value is not large enough.

When dividing, make sure that the latter is relatively large.

Code:

#include <stdio.h>#include <string.h>const int N = 505;typedef long long ll;int n, m;ll max_num, min_num;int books[N];int visit[N];ll Min (const ll a, const ll b) { return a < b ? a : b; } int divide (ll value) {int i = n - 1;int count = 0;ll sum;while (i >= 0) {sum = 0;if (sum + books[i] > value)return m + 1;while (i >= 0 && sum + books[i] <= value) {sum += books[i--];}if (i >= 0)visit[i] = 1;count++;}return count;}int bsearch () {ll left = min_num;ll right = max_num;ll mid;while (left < right) {   mid = left + ((right - left)>>1);   if (divide (mid) <= m)   right = mid;   else   left = mid + 1; }return right;}void solve () {ll ans = bsearch();    memset (visit, 0, sizeof (visit));    int cnt =  divide (ans);for (int i = 0; i < n - 1 && cnt < m; i++) {if (!visit[i]) {visit[i] = 1;cnt++;}}}int main () {int t;scanf ("%d", &t);while (t--) {max_num = 0;scanf ("%d%d", &n, &m);for (int i = 0; i < n; i++) {scanf ("%d", &books[i]);max_num += books[i];if (i == 0)min_num = books[i];elsemin_num = Min(min_num, books[i]);}memset (visit, 0, sizeof (visit));if (m != 1)  solve();for (int i = 0; i < n - 1; i++) {printf ("%d ", books[i]);if (visit[i])printf ("/ ");}printf ("%d\n", books[n - 1]);}return 0;}