c语言函数求利润 c语言利润奖金提成switch
C语言编程,利润提成
#include stdio.h
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#include stdlib.h
void main(void)
{
double money,bouns;
printf("input the money, input a negitive number to leave:");
do{
scanf("%lf",money);
if(money=100000) bouns=money*0.1;
else if(money=200000) bouns=(money-100000)*0.075+100000*0.1;
else if(money=400000) bouns=(money-200000)*0.05+100000*0.075+100000*0.1;
else if(money=600000) bouns=(money-400000)*0.03+200000*0.05+100000*0.075+100000*0.1;
else if(money=1000000) bouns=(money-600000)*0.015+200000*0.03+200000*0.05+100000*0.075+100000*0.1;
else if(money100000) bouns=(money-1000000)*0.01+400000*0.015+200000*0.03+200000*0.05+100000*0.075+100000*0.1;
printf("bouns is %.2lf\n",bouns);
}
while(money=0);
}
#include stdio.h
#include stdlib.h
void main(void)
{
double money,bouns;
int temp;
printf("input the money, input a negitive number to leave:");
do{
scanf("%lf",money);
temp=money/100000;
switch(temp)
{
case 0: bouns=money*0.1; break;
case 1: bouns=(money-100000)*0.075+100000*0.1; break;
case 2:
case 3: bouns=(money-200000)*0.05+100000*0.075+100000*0.1; break;
case 4:
case 5: bouns=(money-400000)*0.03+200000*0.05+100000*0.075+100000*0.1; break;
case 6:
case 7:
case 8:
case 9: bouns=(money-600000)*0.015+200000*0.03+200000*0.05+100000*0.075+100000*0.1; break;
default: bouns=(money-1000000)*0.01+400000*0.015+200000*0.03+200000*0.05+100000*0.075+100000*0.1;
}
C语言求银行本息和利润和
#includestdio.h
void main(){
double num,rate=0.00417,sum=0;
int i;
scanf("%lf",num);
while(num=0){
printf("输入的存款金额必须大于0!\n");
scanf("%lf",num);
}
for(i=1;i=5;i++){
sum=((100+sum)*(1+rate));
}
printf("%.3f\n",sum);
}
C语言编程题 利润提成
#include stdio.h
int main()
{
int z,i;
float sum=0;
printf("input money:");
scanf("%d",i);
if(i=100000)
z=1;
else if(i=200000i100000)
z=2;
else if(i=400000i200000)
z=4;
else if(i=600000i400000)
z=6;
else if(i=1000000i600000)
z=10;
else if(i1000000)
z=11;
switch(z)
{
case 11:sum+=(i-1000000)*0.01;i=1000000;
case 10:sum+=(i-600000)*0.015;i=600000;
case 6:sum+=(i-400000)*0.03;i=400000;
case 4:sum+=(i-200000)*0.05;i=200000;
case 2:sum+=(i-100000)*0.075;i=100000;
case 1:sum+=i*0.1;break;
}
printf("Bonus=%lf",sum);
return 0;
}
调好了 试试
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