HDU 4803 poor warehouse keeper (Greedy)

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The screen displays two values: quantity X and total price y. There are two types of operations, one is to add a total price, to X, 1 + Y; the other is to add a quantity, and the total price will also add a price, to x + 1, Y + Y/X. The total price is an integer after the integer is obtained. The fractional part is ignored. Given a target X and Y, the initial state is. The minimum number of times the target State can be obtained. If the target State cannot be reached, the output is-1.

Idea: if it is to add a total price, the unit price will increase; if it is to add a quantity, the unit price will not change. Therefore, the unit price only increases and does not drop. The number of items must also be changed from 1 to X, that is to say, at least to add X-1 unit price can, then if the unit price too large S, S * (x + 1) ≥y + 1 is definitely not allowed. Therefore, for each I (Quantity), the unit price has an upper limit to ensure that the total price will not overflow when the number is increased.

1 # include <stdio. h> 2 # include <string. h> 3 # include <iostream> 4 5 using namespace STD; 6 7 const double EPS = 1e-9; 8 int main () 9 {10 double X, Y; 11 while (~ Scanf ("% lf", & X, & Y) {12 if (x> Y) 13 {14 printf ("-1 \ n"); 15 continue; 16} 17 double price = (Y + 1-eps)/X; // calculate by the two numbers x and y required in the final question, the final unit price should be S18 double sum = 1.0; 19 int ans = int (x-1); // if the number changes, the total price will not change, so to change to the final X, at least also need to change the X-1 times 20 for (INT I = 1; I <= int (x); I ++) {21 double totalprice = I * price; // take price as the final unit price, then the total price of the current quantity should be 22 int K = (INT) (totalprice-sum ); // how much is the difference to the total price of 23 ans + = K at the unit price; // Add the 24 sum + = K of the difference; // when the number is increased, the total price is also increased by 25 sum = sum * (I + 1) * 1.0/I); 26} 27 printf ("% d \ n ", ans); 28} 29 return 0; 30}

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HDU 4803 poor warehouse keeper (Greedy)