Seven projects, especially within a huaguo mountain is a 100 of, one day for the Monkey King, requires wits with two games respectively,
Xian: (1) all the monkeys into a circle, numbered respectively: 1, 2,... Count, "1" from the beginning to count, to "5" to exit outside, the last one; According to the serial number of the original and then all the monkeys into a big circle, starting from the "1" to count, count to "8" to exit outside, then the rest one,
(2) chan: after a battle of wits left two people in them, his fists, a winner is king,
A thin monkey came late, looking around for a week, to find a place to stand down. In the battle of wits two number off, finally left is his, he is the Monkey King,
Excuse me, how many monkeys are: (1) the mountain of the communist party of China? (2) thin monkey standing position number is how many?

CodePudding user response:

Fyi:

 # include & lt; stdio.h> 

Int game (int * p, int n, int m) 
{
Int I, c, t; 
C=t=0; 
for(i=0; i 
i=0; 
While (t & lt; N - 1) 
{
If (p [I]) c + +; 
If (c==m) 
{
P [I]=0; 
t++; 
C=0; 
} 
I=(I + 1) % n; 
} 

for(i=0; i {
If (p [I]) break; 
} 

The return p [I]; 
} 


Int main (void) 
{
int n; 
Int arr [100]. 

For (n=2; N & lt;=100; N++) 
{
If (game (arr, n, 5)==game (arr, n, 8)) 
{
Printf (" Monkey Count: % d \ \ t t ", n); 
Printf (" Winner ID: \ n \ t % d ", the game (arr, n, 5)); 
} 
} 
return 0; 
}