The classical algorithm of C language--the problem of monkey stealing peach

Title: There are a bunch of peaches on the beach, five monkeys to divide. The first monkey divided this pile of peaches into five copies, one more, and the monkey threw one more into the sea and took a copy. The second monkey to the rest of the peach evenly divided into five parts, and one more, it also threw a lot of one into the sea, took a copy, the third, the fifth monkey is doing this, ask the beach at least how many peaches?

Algorithm Analysis://initial peach number is sum
First time: sum=5*x1+1
Second time: 4*x1=5*x2+1
Third time: 4*x2=5*x3+1
Fourth time: 4*x3=5*x4+1
Fifth time: 4*x4=5*x5+1
Fifth allocation end: The rest 4*x5 a peach, and x5>=1, so i=4*x5>=4. Push j= (I/4) *5+1;i=j each time, after each push end to determine whether j%4 is equal to 0, if 0 continue to push, otherwise update I, until found to be able to complete five push the integer i, finally can get the original number of peaches.

C Language Program:

#include <stdio.h>intMainintAGRC,Char*agrv[]) {    intI, M, J, K, Count;  for(i =4; i<10000; i + =4)/*I assign 5 monkeys to the last remaining peach after the peach, must be a multiple of 4, and then based on this, push up 5 times, if it can be pushed to the fifth time, then the remaining Peach number I is satisfied with the condition, otherwise, the number of remaining peaches to find the minimum number of remaining peaches to meet the conditions*/{Count=0; M=i;  for(k =0; k<5; k++) {J= I/4*5+1; I=J; if(J%4==0) Count++; Else                 Break; } I=m; if(Count = =4) {printf ("the original minimum number of Peaches is:%d\n", J);  Break; }} system ("Pause"); return 0;}

The classical algorithm of C language--the problem of monkey stealing peach