Ultraviolet A 812-Trade on Verweggistan (violent)

Question connection: Ultraviolet A 812-Trade on Verweggistan

N sequences are given. Each sequence is given k, indicating that there are k elements. Each row can only be retrieved from the beginning, taking I (I ≤ k ), now we need to remove c [I] before each sequence, ensure that sum = Σ | 10-num | (num is the set of the selected number) is the maximum, and output the number that can be selected, if the number is greater than 10, the minimum output value is 10.

Solution: traverse each sequence once to find out the maximum value and the number that can constitute the maximum value. Note that you can take 0 and then recursively solve all composition conditions.

#include 
 
  #include 
  
   #include 
   
    #include using namespace std;const int tmp = 10;const int N = 60;const int M = 1005;const int INF = 0x3f3f3f3f;int n, ans, cnt[M];vector
    
      v[N];void init() {ans = 0;memset(cnt, 0, sizeof(cnt));int m, num;for (int i = 0; i < n; i++) {scanf("%d", &m); v[i].clear();int sum = 0, Max = 0;v[i].push_back(0);for (int j = 0; j < m; j++) {scanf("%d", &num);sum += (tmp - num);if (sum > Max) {v[i].clear();v[i].push_back(j+1);Max = sum;} else if (sum == Max) {v[i].push_back(j+1);}}ans += Max;}}void dfs(int d, int s) {if (d >= n) {cnt[s] = 1;return ;}int top = v[d].size();for (int i = 0; i < top; i++) {dfs(d + 1, s + v[d][i]);}}void solve() {dfs(0, 0);int c = 0;for (int i = 0; i < M; i++) {if (c >= tmp) break;if (cnt[i]) {printf(" %d", i);c++;}}printf("\n");}int main() {int cas = 0;while (scanf("%d", &n) == 1 && n) {init();if (cas) printf("\n");printf("Workyards %d\n", ++cas);printf("Maximum profit is %d.\n", ans);printf("Number of pruls to buy:");solve();}return 0;}