标签：UNC ado blog return size http 情况 process one
当下，坐公交或者地铁时大部分人都是刷卡的。不过，时至今日还在用现金支付的人还是比想象的多。本题我们以安置在公交上的零钱兑换机为背景。
package main
import "fmt"
const (
coin10 = 10
coin50 = 50
coin100 = 100
coin500 = 500
)
func mostCount(money, coinDeno int)int{
mostC := money / coinDeno
if mostC > 15{
return 15
}else{
return mostC
}
}
func main(){
money := 1000
coin10MostCount := mostCount(money, coin10)
coin50MostCount := mostCount(money, coin50)
coin100MostCount := mostCount(money, coin100)
coin500MostCount := mostCount(money, coin500)
n := 0
for a:=0;a<=coin500MostCount;a++ {
for b:=0;b<=coin100MostCount;b++{
for c:=0;c<=coin50MostCount;c++{
for d:=0;d<=coin10MostCount;d++{
if 500*a + 100*b + 50*c + 10*d == money && a + b + c + d <= 15{
fmt.Printf("%d = 500*%2d + 100*%2d + 50*%2d + 10*%2d\n", money, a, b, c, d)
n++
}
}
}
}
}
fmt.Println("共", n, "种组合")
}结果：
1000 = 500* 0 + 100* 5 + 50*10 + 10* 0
1000 = 500* 0 + 100* 6 + 50* 8 + 10* 0
1000 = 500* 0 + 100* 7 + 50* 6 + 10* 0
1000 = 500* 0 + 100* 8 + 50* 4 + 10* 0
1000 = 500* 0 + 100* 9 + 50* 1 + 10* 5
1000 = 500* 0 + 100* 9 + 50* 2 + 10* 0
1000 = 500* 0 + 100*10 + 50* 0 + 10* 0
1000 = 500* 1 + 100* 0 + 50* 9 + 10* 5
1000 = 500* 1 + 100* 0 + 50*10 + 10* 0
1000 = 500* 1 + 100* 1 + 50* 7 + 10* 5
1000 = 500* 1 + 100* 1 + 50* 8 + 10* 0
1000 = 500* 1 + 100* 2 + 50* 5 + 10* 5
1000 = 500* 1 + 100* 2 + 50* 6 + 10* 0
1000 = 500* 1 + 100* 3 + 50* 3 + 10* 5
1000 = 500* 1 + 100* 3 + 50* 4 + 10* 0
1000 = 500* 1 + 100* 4 + 50* 0 + 10*10
1000 = 500* 1 + 100* 4 + 50* 1 + 10* 5
1000 = 500* 1 + 100* 4 + 50* 2 + 10* 0
1000 = 500* 1 + 100* 5 + 50* 0 + 10* 0
1000 = 500* 2 + 100* 0 + 50* 0 + 10* 0
共 20 种组合貌似复杂，做起来其实不难，把各种情况都让计算机试一遍就好了。
标签：UNC ado blog return size http 情况 process one
原文地址：http://blog.51cto.com/johnnyloo/2136281
