IntType gcd( const IntType & m, const IntType & n )
{
if( m > 0 )
return gcd1( n, m );
else
return gcd1( n, -m );
}
/* integers to compute the GCD of */
static long gcd1(long p, long q) {
if(q > p) return gcd1(q,p);
if(q == 0) return p;
printf(" gcd(%ld,%ld) &=& gcd(%ld \\mod %ld, %ld) = gcd(%ld,%ld) \n",
p, q, p, q, q, q, p%q);
return( gcd1(q, p % q) );
}
int main() {
int a, b;
printf("Enter two number: \n");
while (scanf("%d%d", &a, &b) == 2) {
if(a == 0 || b == 0)
break;
printf("gcd1: %d\n", gcd1(a, b));
printf("gcd2: %d\n", gcd1(a, b));
printf("gcd3: %d\n", gcd1(a, b));
}
return 0;
}
/*********************************************
* ch5, p106, 5. 输入两个正整数m和n，求其最大公约数和最小公倍数
*********************************************/
void ch5_5()
{
printf("====ch5_5()\n");
int m,n,gcd;
printf("请输入两个正整数，求其最大公约数和最小公倍数。\n");
scanf("%d%d",&m,&n);
gcd = gcd1(m,n);
printf("%d,%d的最大公约数=%d,",m,n,gcd);
printf("最小公倍数=%d\n",m*n/gcd);
/********************************************
以下为各种方法的验证
// 循环语句的辗除法,求两个整数的最大公约数
printf("319,377最大公约数：%d\n",gcd1(319,377)); //29
// 循环语句的辗除法,求两个整数的最大公约数
printf("319,377最大公约数：%d\n",gcd2(319,377)); //29
// 递归调用的辗除法,求两个整数的最大公约数
printf("377,319最大公约数：%d\n",gcd3(377,319)); //29
printf("319,377最大公约数：%d\n",gcd3(319,377)); //29
printf("377,319最大公约数：%d\n",gcd4(377,319)); //29
printf("319,377最大公约数：%d\n",gcd4(319,377)); //29
printf("377,319最大公约数：%d\n",gcd5(377,319)); //29
printf("319,377最大公约数：%d\n",gcd5(319,377)); //29
// 最小公倍数=两整数的乘积÷最大公约数
printf("319,377最小公倍数：%d\n",multiple (319,377));
**********************************************/
}
long gcd(long m)
{
long i=2;
while(gcd1(i,m)!=1)
i++;
return i;
}
// n is guaranteed non-negative
IntType gcd1( const IntType & n, const IntType & m )
{
if( n % m == 0 )
return m;
else
return gcd1( m, n % m );
}
int main(void) {
long p, q;
long x, y, g1, g2;
while(scanf("%ld %ld", &p, &q)!=EOF) {
printf("gcd of p = %ld and q = %ld = %ld\n",p, q, g1 = gcd1(p, q));
printf(" %ld*%ld + %ld*%ld = %ld\n",p, x, q, y, g2 = gcd(p, q, &x, &y));
if(g1 != g2) printf("ERROR: GCD\n");
if((p*x + q*y) != g1) printf("ERROR: DIOPHONINE SOLUTION WRONG!\n");
}
return 0;
}
int main(void)
{
int gcd, lcm, m, n;
int repeat, ri;
scanf("%d", &repeat);
for(ri = 1; ri <= repeat; ri++){
scanf("%d", &m);
scanf("%d", &n);
if(m <= 0 || n <= 0)
printf("m <= 0 or n <= 0");
else{
/*---------*/
gcd=gcd1(m,n);
lcm=m/gcd*n;
printf("%d is the least common multiple of %d and %d, %d is the greatest common divisor of %d and %d.\n", lcm, m, n, gcd, m, n);
}
}
}
void ch1()
{
// 循环语句的辗除法,求两个整数的最大公约数
printf("319,377最大公约数：%d\n",gcd1(319,377)); //29
// 循环语句的辗除法,求两个整数的最大公约数
printf("319,377最大公约数：%d\n",gcd2(319,377)); //29
// 递归调用的辗除法,求两个整数的最大公约数
printf("377,319最大公约数：%d\n",gcd3(377,319)); //29
printf("319,377最大公约数：%d\n",gcd3(319,377)); //29
printf("377,319最大公约数：%d\n",gcd4(377,319)); //29
printf("319,377最大公约数：%d\n",gcd4(319,377)); //29
printf("377,319最大公约数：%d\n",gcd5(377,319)); //29
printf("319,377最大公约数：%d\n",gcd5(319,377)); //29
// 最小公倍数=两整数的乘积÷最大公约数
//printf("319,377最小公倍数：%d\n",multiple (319,377));
}
int lcm(int a, int b)
{
if (0 == b)return 0;
return a*gcd1(a, b) / b;
}
int gcd1(int a, int b)
{
return b == 0 ? 0 : gcd1(b, a % b);
}
inline bigInt gcd(const bigInt& a, const bigInt& b)
{
return gcd1(abs(a), abs(b));
}
inline bigInt gcd1(const bigInt& a, const bigInt& b)
{
if (a == 0)
return b;
return gcd1(b % a, a);
}
// 最小公倍数=两整数的乘积/最大公约数
int multiple (int a,int b)
{
printf("最小公倍数=两整数的乘积÷最大公约数\n");
return(a*b/gcd1(a,b));
}
int gcd1(int a,int b)
{
if (b==0) return a;
else return gcd1(b,a%b);
}
