已解决问题
谷歌丹甫欧尼用户在2015.12.07提交了关于“蒙面唱将猜猜猜C++两个函数值返回同一个主函数，其中一个函数在用时还用到了第一个函数的结果”的提问，欢迎大家涌跃发表自己的观点。目前共有2个回答，最后更新于2025-03-01T15:14:13。#include<stdio.h>
#include<string>
doubletax(intsalary);
doublebumanzu(doubleb)
{
doubled;
inti;
inta[9][3]={{7800,3000,4000},{7000,4000,5000},{28000,16000,7000},{15000,6000,10000},{12000,5000,4000},{4300,1500,0},{6500,3400,3400},{5900,2000,2400},{2900,1000,0}};
charname[9][10]={"Tom","Kate","Weihe","Xiaoming","Zhangdad","Sunnan","Li**n","Wanghan","Me"};
for(i=0;i<9;i++)
{
d=tax(a[i][0]);
b=a[i][0]-d-a[i][1]-a[i][2];
}
return(b);
}
doubletax(intsalary)
{
intk;
doubleq,m,n;
k=salary-3500;
if(k>0&&k<=1500)
{
m=0.03;
n=0;
q=k*m-n;
}
if(k<0)
{
q=0;
}
if(k>1500&&k<=4500)
{
m=0.1;
n=105;
q=k*m-n;
}
if(k>4500&&k<=9000)
{
m=0.2;
n=555;
q=k*m-n;
}
if(k>9000&&k<=35000)
{
m=0.25;
n=1005;
q=k*m-n;
}
if(k>35000&&k<=55000)
{
m=0.3;
n=2755;
q=k*m-n;
}
if(k>55000&&k<=80000)
{
m=0.35;
n=5505;
q=k*m-n;
}
if(k>80000)
{
m=0.45;
n=13505;
q=k*m-n;
}
return(q);
}
voidmain()
{
charname[9][10]={"Tom","Kate","Weihe","Xiaoming","Zhangdad","Sunnan","Li**n","Wanghan","Me"};
inta[9][3]={{7800,3000,4000},{7000,4000,5000},{28000,16000,7000},{15000,6000,10000},{12000,5000,4000},{4300,1500,0},{6500,3400,3400},{5900,2000,2400},{2900,1000,0}};
intsum;
sum=0;
doublem,q,n;
inti,c,u;
for(i=0;i<9;i++)
{
m=tax(a[i][0]);
n=a[i][0]-m;
sum=sum+n;
printf("%s的税后收入值为%f\n",name[i],n);
}
q=sum/9;
for(i=0;i<9;i++)
{
c=bumanzu(i);
u=a[i][1]+a[i][2];
if(c<0)
{
printf("%s的收入不能满足支出，收入为%d,支出为%d\n",name[i],a[i][0],u);
}
}
printf("平均税后收入为%f\n",q);
}
详细问题描述及疑问：
#include<stdio.h>
#include<string>
doubletax(intsalary);
doublebumanzu(doubleb)
{
doubled;
inti;
inta[9][3]={{7800,3000,4000},{7000,4000,5000},{28000,16000,7000},{15000,6000,10000},{12000,5000,4000},{4300,1500,0},{6500,3400,3400},{5900,2000,2400},{2900,1000,0}};
charname[9][10]={"Tom","Kate","Weihe","Xiaoming","Zhangdad","Sunnan","Li**n","Wanghan","Me"};
for(i=0;i<9;i++)
{
d=tax(a[i][0]);
b=a[i][0]-d-a[i][1]-a[i][2];
}
return(b);
}
doubletax(intsalary)
{
intk;
doubleq,m,n;
k=salary-3500;
if(k>0&&k<=1500)
{
m=0.03;
n=0;
q=k*m-n;
}
if(k<0)
{
q=0;
}
if(k>1500&&k<=4500)
{
m=0.1;
n=105;
q=k*m-n;
}
if(k>4500&&k<=9000)
{
m=0.2;
n=555;
q=k*m-n;
}
if(k>9000&&k<=35000)
{
m=0.25;
n=1005;
q=k*m-n;
}
if(k>35000&&k<=55000)
{
m=0.3;
n=2755;
q=k*m-n;
}
if(k>55000&&k<=80000)
{
m=0.35;
n=5505;
q=k*m-n;
}
if(k>80000)
{
m=0.45;
n=13505;
q=k*m-n;
}
return(q);
}
voidmain()
{
charname[9][10]={"Tom","Kate","Weihe","Xiaoming","Zhangdad","Sunnan","Li**n","Wanghan","Me"};
inta[9][3]={{7800,3000,4000},{7000,4000,5000},{28000,16000,7000},{15000,6000,10000},{12000,5000,4000},{4300,1500,0},{6500,3400,3400},{5900,2000,2400},{2900,1000,0}};
intsum;
sum=0;
doublem,q,n;
inti,c,u;
for(i=0;i<9;i++)
{
m=tax(a[i][0]);
n=a[i][0]-m;
sum=sum+n;
printf("%s的税后收入值为%f\n",name[i],n);
}
q=sum/9;
for(i=0;i<9;i++)
{
c=bumanzu(i);
u=a[i][1]+a[i][2];
if(c<0)
{
printf("%s的收入不能满足支出，收入为%d,支出为%d\n",name[i],a[i][0],u);
}
}
printf("平均税后收入为%f\n",q);
}