(温故而知新) 最小公倍数最大公约数的 C实现

看到书里面实现最大公约数的代码特别长，心里面难受，就自个儿动手了，只需一行代码就可以搞定。

于是一发不可收拾，用了两个算法。后来又补充了一下最小公倍数，不调用GCD，也给出了两个算法：

 

#include<stdio.h> int gcd1(int m,int n){int t; while(m!=0) t=m,m=n%m,n=t;} int gcd2(int m,int n){while(m-n) if(m>n) m-=n;else n-=m;} int lcm1(int a,int b){int m=a,n=b,t; while(m!=0) t=m,m=n%m,n=t;return a/n*b;} int lcm2(int a,int b){int m=a,n=b; while(m-n) if(m>n) n+=b;else m+=a;} int main(void){ int m,n; printf("please input two positive integers:/n"); scanf("%d%d",&m,&n); printf("GCD1(%d,%d)=/t%d/n",m,n,gcd1(m,n)); printf("GCD2(%d,%d)=/t%d/n",m,n,gcd2(m,n)); printf("LCM1(%d,%d)=/t%d/n",m,n,lcm1(m,n)); printf("LCM2(%d,%d)=/t%d/n",m,n,lcm2(m,n)); return 0; }

 

以上代码在GCC里面，编译通过，但添加-Wall选项，则有警告信息，

gcd.c: In function 'lcm2':
gcd.c:5:1: warning: control reaches end of non-void function
gcd.c: In function 'gcd2':
gcd.c:3:1: warning: control reaches end of non-void function
gcd.c: In function 'gcd1':
gcd.c:2:1: warning: control reaches end of non-void function

 

呵呵，很明显，因为我上面三个函数没有显式的 return语句。

 

当然了，要在VS2008里面编译通过，稍微改改就是了：

 

#define _CRT_SECURE_NO_WARNINGS #include<stdio.h> #include<stdlib.h> int gcd1(int m,int n){int t; while(m!=0) t=m,m=n%m,n=t;return n;} int gcd2(int m,int n){while(m-n) if(m>n) m-=n;else n-=m;return n;} int lcm1(int a,int b){int m=a,n=b,t; while(m!=0) t=m,m=n%m,n=t;return a/n*b;} int lcm2(int a,int b){int m=a,n=b; while(m-n) if(m>n) n+=b;else m+=a;return m;} int main(void){ int m,n; printf("please input two positive integers:/n"); scanf("%d%d",&m,&n); printf("GCD1(%d,%d)=/t%d/t",m,n,gcd1(m,n)); printf("GCD2(%d,%d)=/t%d/n",m,n,gcd2(m,n)); printf("LCM1(%d,%d)=/t%d/t",m,n,lcm1(m,n)); printf("LCM2(%d,%d)=/t%d/n/n",m,n,lcm2(m,n)); system("date/t&&pause"); return EXIT_SUCCESS; }