1 条题解
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2lichengjun LV 10 @ 2023-1-12 12:09:47
阿巴阿巴……
#include<iostream> #include<iomanip> #include<stdio.h> #include<math.h> #include<string> #include<cstring> using namespace std; double n,tc,sum=0;//利润,提成,奖金 void js(double n,double tc,double bf1,double bf2){//bf1和bf2为判断部分 if(n>bf1){ if(n<bf2)sum+=(n-bf1)*tc; else sum+=(bf2-bf1)*tc; }else return; } int main(){ cin>>n; js(n,0.1,0,100000); js(n,0.075,100000,200000); js(n,0.05,200000,400000); js(n,0.03,400000,600000); js(n,0.015,600000,1000000); js(n,0.01,1000000,pow(10,10));//这里的bf2没有上限,pow(a,b)=a的b次方 printf("%.2lf",sum); return 0; }
- 1
信息
- ID
- 860
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 6
- 标签
- 递交数
- 59
- 已通过
- 20
- 上传者