1 条题解
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0
C :
#include<stdio.h> int main() { int pro,money,i; scanf("%d",&pro); if(pro<=100000) money=pro*0.1; else if(100000<pro&&pro<=200000) money=100000*0.1+(pro-100000)*0.075; else if(200000<pro&&pro<=400000) money=100000*0.1+100000*0.075+(pro-200000)*0.05; else if(400000<pro<=600000) money=100000*0.1+100000*0.075+200000*0.05+(pro-400000)*0.03; else if(600000<pro<=1000000) money=100000*0.1+100000*0.075+200000*0.05+200000*0.03+(pro-600000)*0.015; else money=100000*0.1+100000*0.075+200000*0.05+200000*0.03+400000*0.015+(pro-1000000)*0.01; printf("%d\n",money); return 0; }
C++ :
#include<iostream> #include<iostream> using namespace std; int main() {int a;cin>>a;float d; if(a<=100000)d=a*0.1; else if(a<=200000)d=10000+(a-100000)*0.075; else if(a<=400000)d=10000+7500+(a-200000)*0.05; else if(a<=600000)d=27500+(a-400000)*0.03; else if(a<=1000000)d=33500+(a-600000)*0015; else d+=39500+(d-1000000)*0.01; cout<<int (d)<<endl;return 0;}
Pascal :
var n:longint; x:real; begin readln(n); if n<=100000 then x:=n*0.1; if (n>100000)and(n<=200000) then x:=n*0.1; if (n>200000)and(n<=400000) then x:=n*0.05; if (n>400000)and(n<=600000) then x:=n*0.03; if (n>600000)and(n<=1000000) then x:=n*0.015; if n>1000000 then x:=n*0.01; writeln(x:0:0); end.
- 1
信息
- ID
- 1417
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- (无)
- 标签
- (无)
- 递交数
- 0
- 已通过
- 0
- 上传者