摘要:
PAT B1001害死人不偿命的(3n+1) 猜想 #include<stdio.h> int main() { int n,countnum=0; scanf("%d",&n); while(n!=1) { if(n%2==0) { n=n/2; } else { n=(3*n+1)/2; } c 阅读全文
摘要:
PAT B1001害死人不偿命的(3n+1) 猜想 #include<stdio.h> int main() { int n,countnum=0; scanf("%d",&n); while(n!=1) { if(n%2==0) { n=n/2; } else { n=(3*n+1)/2; } c 阅读全文
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