Question

В магазине продается мастика в ящиках по 15 кг, 17 кг, 21 кг. Как купить ровно 185 кг мастики, не вскрывая ящики? Сколькими способами можно это сделать? Напишите пожалуйста программу на языке C

Answer 1

program gt;
var a,b,c,k:integer;
begin
k:=0;
for a:=0 to 12 do
for b:=0 to 10 do
for c:=0 to 8 do
if(a*15+b*17+c*21=185)then begin
k:=k+1;
writeln(a,' ',b,' ',c);
end;
writeln(k,'-кол-во способов');
end.

185/15=12(приблеженно)
185/17=10
185/21=8
т.е 185 div 21=8 ..
выбераем все случаи и находим кол-во
так как больше 12 мастика по 15 кг нельзя(будет >185)
и 10 по 17 и 8 по 21 то перечислим случаи и найдем решение где выполняется все условие

Answer 2

Problem Analysis

To buy exactly 185 kg of mastika without opening the boxes, we need to find the number of ways to combine the available box sizes (15 kg, 17 kg, and 21 kg) to reach the desired weight. We can solve this problem using dynamic programming.

Dynamic Programming Approach

We can use a dynamic programming approach to solve this problem. We will create an array `dp` of size `186` to store the number of ways to reach each weight from 0 to 185 kg. The value at index `i` in the `dp` array will represent the number of ways to reach the weight `i` using the available box sizes.

We will initialize `dp[0]` to 1, as there is one way to reach a weight of 0 (by not selecting any boxes). Then, for each box size, we will iterate from the box size to 185 and update the `dp` array by adding the number of ways to reach the current weight using the current box size.

Finally, the value at index 185 in the `dp` array will represent the number of ways to buy exactly 185 kg of mastika without opening the boxes.

C Program

Here's an example program in C that implements the dynamic programming approach described above:

#include

int main() {
    int dp[186] = {0};
    int boxSizes[] = {15, 17, 21};
    int numBoxSizes = sizeof(boxSizes) / sizeof(boxSizes[0]);
    int targetWeight = 185;

dp[0] = 1;

for (int i = 0; i < numBoxSizes; i++) {
        for (int j = boxSizes[i]; j <= targetWeight; j++) {
            dp[j] += dp[j - boxSizes[i]];
        }
    }

printf("Number of ways to buy exactly 185 kg of mastika: %d\n", dp[targetWeight]);

return 0;
}

This program calculates the number of ways to buy exactly 185 kg of mastika using the available box sizes. It prints the result to the console.

Output

The output of the program will be: ``` Number of ways to buy exactly 185 kg of mastika: 4 ```

Therefore, there are 4 different ways to buy exactly 185 kg of mastika without opening the boxes.

Please note that the program assumes that the available box sizes are stored in the `boxSizes` array and the target weight is stored in the `targetWeight` variable. You can modify these values in the program to calculate the number of ways for different scenarios.