题目：输入两个正整数m和n，求其最大公约数和最小公倍数
分析：使用欧几里德算法
package main
import (
"fmt"
)
/*最大公约数*/
func euclidean(a, b int) int {
for b != 0 {
a, b = b, a%b
}
return a
}
/*最小公倍数*/
func leastcommmultiple(a, b, c int) int {
return (a / c) * (b / c) * c
}
func main() {
var m, n, k int
fmt.Scanf("%d %d", &m, &n)
k = euclidean(m, n)
fmt.Printf("%d和%d的最大公约数是%d\n", m, n, k)
fmt.Printf("%d和%d的最小公倍数是%d\n", m, n, leastcommmultiple(m, n, k))
}
运行结果：
48 95
48和95的最大公约数是1
48和95的最小公倍数是4560
...
40 91
40和91的最大公约数是1
40和91的最小公倍数是3640
...
324 958
324和958的最大公约数是2
324和958的最小公倍数是155196
package main
import (
"fmt"
)
/*最大公约数*/
func euclidean(a, b int) int {
for b != 0 {
a, b = b, a%b
}
return a
}
/*最小公倍数*/
func leastcommmultiple(a, b, c int) int {
return (a / c) * (b / c) * c
}
func main() {
var m, n, k int
fmt.Scanf("%d %d", &m, &n)
k = euclidean(m, n)
fmt.Printf("%d和%d的最大公约数是%d\n", m, n, k)
fmt.Printf("%d和%d的最小公倍数是%d\n", m, n, leastcommmultiple(m, n, k))
}
package main
import (
"fmt"
)
/*最大公约数*/
func euclidean(a, b int) int {
for b != 0 {
a, b = b, a%b
}
return a
}
/*最小公倍数*/
func leastcommmultiple(a, b, c int) int {
return (a / c) * (b / c) * c
}
func main() {
var m, n, k int
fmt.Scanf("%d %d", &m, &n)
k = euclidean(m, n)
fmt.Printf("%d和%d的最大公约数是%d\n", m, n, k)
fmt.Printf("%d和%d的最小公倍数是%d\n", m, n, leastcommmultiple(m, n, k))
}
package main
import (
"fmt"
)
/*最大公约数*/
func euclidean(a, b int) int {
for b != 0 {
a, b = b, a%b
}
return a
}
/*最小公倍数*/
func leastcommmultiple(a, b, c int) int {
return (a / c) * (b / c) * c
}
func main() {
var m, n, k int
fmt.Scanf("%d %d", &m, &n)
k = euclidean(m, n)
fmt.Printf("%d和%d的最大公约数是%d\n", m, n, k)
fmt.Printf("%d和%d的最小公倍数是%d\n", m, n, leastcommmultiple(m, n, k))
}
package main
import (
"fmt"
)
/*最大公约数*/
func euclidean(a, b int) int {
for b != 0 {
a, b = b, a%b
}
return a
}
/*最小公倍数*/
func leastcommmultiple(a, b, c int) int {
return (a / c) * (b / c) * c
}
func main() {
var m, n, k int
fmt.Scanf("%d %d", &m, &n)
k = euclidean(m, n)
fmt.Printf("%d和%d的最大公约数是%d\n", m, n, k)
fmt.Printf("%d和%d的最小公倍数是%d\n", m, n, leastcommmultiple(m, n, k))
}
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