找循环位数是奇数的数有多少个
这个自己很难写出来，完全不能暴力
*链接 *上面说的很好，上面的算法实现就好了。
就是上面的
Java程序：
package project61;
public class P64{
void run(){
int count = 0;
int m = 0;
int d = 1;
int a0 = 0;
int a = 0;
int period = 0;
for(int S = 2;S<10000;S++){
period = 0;
m = 0;
d = 1;
a0 = (int) (Math.sqrt(S));
if(a0*a0 == S) continue;
a = a0;
do{
m = d*a - m;
d = (S-m*m)/d;
a = (a0+m)/d;
period++;
}while(a!=2*a0);
if(period%2==1) count++;
}
System.out.println(count);
}
public static void main(String[] args){
long start = System.currentTimeMillis();
new P64().run();
long end = System.currentTimeMillis();
long time =end - start;
System.out.println("run time:"+ time/1000+"s"+time%1000+"ms");
}
}
Python程序
import time as time start = time.time() count = 0 for S in range(2,10000):
m = 0
d = 1
a0 = int(S**0.5)
if a0*a0 == S :continue
preiod = 0
a= a0
while a!=2*a0:
m = d*a - m
d = (S - m*m)/d
a = int((a0 + m)/d)
preiod+=1
if preiod%2==1:count +=1 end = time.time()
print "time={0} secs,count={1}".format((end-start),count)
上面两个程序几乎一样的