Sale of goldfish

Sale of goldfish
The buyer and seller sold the goldfish in a tank for five times. The last time the system sold half of the whole, it added 1/2; the second time it sold the remaining 1/3 and 1/3; for the third time, the remaining 1/4 and 1/4 items are sold. for the fourth time, the remaining 1/5 and 1/5 items are sold. For the last time, the remaining 11 items are sold. How many goldfish are there in the original fish tank?
* Question analysis and Algorithm Design
All the fish in the question are sold for five times, and the strategy for each sale is the same; the remaining (J + 1) points of the second (J + 1) are repeatedly added with 1/(J + 1. In the fifth round, we sold all the remaining 11 items for the fourth round.
Assuming that the total number of times J fish is X, the J fish is left:
X-(x + 1)/(J + 1)
When the fourth sale is complete, there should be 11 remaining items. If X meets the preceding requirements, X is the solution of the question.
Note that "(x + 1)/(J + 1)" must meet the division criteria. The initial value of test X can start from 23, and the test step is 2, because the value of X must be an odd number.
* Program Description and comment
# Include <stdio. h>
Void main ()
{
Int I, j, n = 0, x;/* n is the flag variable */
For (I = 23; n = 0; I + = 2)/* control the test step and process */
{
For (j = 1, x = I; j <= 4 & x> = 11; j ++)/* four sales operations */
If (x + 1) % (J + 1) = 0)/* If the Division conditions are met, the actual sales operation is performed */
X-= (x + 1)/(J + 1 );
Else {x = 0; break;}/* otherwise, the computing process is stopped */
If (j = 5 & X = 11)/* If the remaining 11 items in the fourth time meet the question */
{
Printf ("there are % d fishes at first. N", I);/* output result */
N = 1;/* control the exit test process */
}
}
}
* Running result
There are 59 fishes at first.