Ultraviolet A: 10670-Work ction (Greedy)

Ultraviolet A: 10670-Work ction (Greedy)

Question: 10670-Work ction

N, m, L. n represents the total number of paperworks given by the boss, m represents the remaining number, and L represents the minimum cost you need from so many companies.

Solution:

1. L branch companies. Each line is named by the company: A, B; A represents A copy of the money required by paperwork, B Represents helping you reduce the total number of paperworks to half the money required for the call. Note that you have paperworks on hand instead of the number of paperworks that the boss requires you to complete. It has been stuck here for a long time. Otherwise, the final quantity cannot be smaller than m. Because the boss requires m copies to be left.

2. Separate the input string from the name and Money A and B.

3. Calculate the sorting output of the amount.

Code:

# Include
 
  
# Include
  
   
Using namespace std; const int N = 105; int n, m, l; struct MONEY {char name [N]; int money; int a, B;} M [N]; char str [N]; int cmp (const MONEY & x, const MONEY & y) {if (x. money <y. money) return true; else if (x. money> y. money) return false; else return strcmp (x. name, y. name) <0? True: false;} // separated string names and charges void handle (int j) {int k = 0; for (int I = 0; I <strlen (str ); I ++) {if (str [I]! = ':') K ++; elsebreak;} memcpy (M [j]. name, str, sizeof (str); M [j]. name [k] = '\ 0'; int sum = 0; for (int I = k + 1; I <strlen (str); I ++) {if (str [I]! = ',') Sum = sum * 10 + str [I]-'0'; else {M [j]. a = sum; sum = 0 ;}} M [j]. B = sum;} // calculation fee void cal (int j) {int num = n; int sum = 0; while (1) {if (num/2> = m & (num-num/2) * M [j]. a> = M [j]. b) {sum + = M [j]. b; num/= 2;} else {sum + = (num-m) * M [j]. a; M [j]. money = sum; break ;}}int main () {int t; char ch; scanf ("% d", & t); for (int I = 1; I <= t; I ++) {printf ("Case % d \ n", I); scanf ("% d % c", & n, & m, & l, & ch); for (int j = 0; j <l; j ++) {gets (str); handle (j ); cal (j);} sort (M, M + l, cmp); for (int j = 0; j <l; j ++) printf ("% s % d \ n", M [j]. name, M [j]. money);} return 0 ;}