雷火笔试第二题到底怎么做？
第二题换了两种语言都是43%通过率
下面是我的Go和Python代码
package main
import (
"fmt"
"sort"
)
func solution2(Hp, LHp, UHp, N int, attacks []int) int {
result := 0
sort.Sort(sort.Reverse(sort.IntSlice(attacks)))
//For DEBUG
//path := make([]int, 0)
//defer func() {
// fmt.Println(path)
//}()
for _, atk := range attacks {
for {
atkedHp := Hp - atk
if atkedHp > UHp {
//For DEBUG
//path = append(path, atk)
result++
Hp -= atk
} else if LHp <= atkedHp && atkedHp <= UHp {
result++
//For DEBUG
//path = append(path, atk)
return result
} else if atkedHp < LHp {
break
}
}
}
return 0
}
func main() {
M := 0
fmt.Scan(&M)
for m := 0; m < M; m++ {
Hp, LHp, UHp, N := 0, 0, 0, 0
fmt.Scan(&Hp, &LHp, &UHp, &N)
attacks := make([]int, N)
for i := range attacks {
atk := 0
fmt.Scan(&atk)
attacks[i] = atk
}
fmt.Println(solution2(Hp, LHp, UHp, N, attacks))
}
} # -*-coding:utf-8-*- # coding=utf-8 import sys def solution2( Hp, LHp, UHp, attacks): result = 0 attacks.sort(reverse=True) for atk in attacks: while True: atkedHp = Hp - atk if atkedHp > UHp: result += 1 Hp -= atk elif LHp <= atkedHp and atkedHp <= UHp: result += 1 return result elif atkedHp < LHp: break return 0 if __name__ == "__main__": # 读取第一行的n n = int(sys.stdin.readline().strip()) for i in range(n): # 读取每一行 line = sys.stdin.readline().strip() # 把每一行的数字分隔后转化成int列表 values = list(map(int, line.split())) Hp, LHp, UHp, N = values[0], values[1], values[2], values[3] line = sys.stdin.readline().strip() # 把每一行的数字分隔后转化成int列表 attacks = list(map(int, line.split())) print(solution2( Hp, LHp, UHp, attacks))
