UVA 812 (violence)

Test instructions: There are n shelves of B goods, and then from left to right to give the price of the goods on each shelf, to be sold at a flat price of 10, ask if the Buy form is a stack, to buy a goods must be his left to buy things to be able to ask the maximum profit is how much, to buy how many goods to achieve maximum profit, Output all possible conditions (up to 10) in dictionary order.

Problem: Water, violence out of all circumstances.

Example of the second set of samples:

3 10 9) 10 10

Number of purchases possible: 2 4 5
9 17 24 30 30 24 24 30 24

Number of purchases possible: 4 5 8

Then there are 6 numbers of permutations combined to get all the solutions. Note the number may be 0.

#include <stdio.h> #include <string.h> #include <algorithm>using namespace std;const int M = 1005;const int N = 55;int Box[n], N, Num[n][n], mp[m], num2[n];void dfs (int cur, int sum) {if (cur >= N) {mp[sum] = 1;return;} for (int i = 0; i < num2[cur]; i++) DFS (cur + 1, sum + num[cur][i]);} int main () {int cas = 0;while (scanf ("%d", &n) = = 1 && N) {memset (MP, 0, sizeof (MP)), int temp, res = 0;for (in t i = 0; I < n;  i++) {scanf ("%d", &temp), int sum = 0, Maxx = 0;for (int j = 0; J < temp; J + +) {scanf ("%d", &box[j]); sum + = 10- Box[j];maxx = max (Maxx, sum);} Res + = Maxx;sum = 0;int cnt = 0;if (Maxx = = 0) num[i][cnt++] = 0;for (int j = 0; J < temp; J + +) {sum + = 10-box[j];if ( sum = = Maxx) num[i][cnt++] = j + 1;} Num2[i] = cnt;} if (CAS) printf ("\ n");p rintf ("Workyards%d\n", ++cas);p rintf ("Maximum profit is%d.\n", res);p rintf ("Number of the pruls to BU Y: ");d FS (0, 0); temp = 0;for (int i = 0; i < M; i++) {if (temp = =) break;if (Mp[i]) {temp++;p rintf (" %d ", i);}} printf ("\ n");} return 0;}

UVA 812 (violence)