0316复利计算

#include<stdio.h>
#include<math.h>
void Jisuan();
double p,i,f;
int n;
void Danli()
{
printf("请输入本金：");
scanf("%lf",&p);
printf("请输入年利率（百分比）：");
scanf("%lf",&i);
printf("请输入存储年限：");
scanf("%d",&n);
f = p+p*(i/100)*n;
printf("单利终值为：%0.2lf\n",f);
}

void Fuli()
{
printf("请输入本金：");
scanf("%lf",&p);
printf("请输入年利率（百分比）：");
scanf("%lf",&i);
printf("请输入存储年限：");
scanf("%d",&n);
f = p*(pow(1+i/100,n));
printf("复利终值为：%0.2lf\n",f);
}

void Need()
{
printf("请输入目标金额：");
scanf("%lf",&f);
printf("请输入目标利率：");
scanf("%lf",&i);
printf("请输入目标年限：");
scanf("%d",&n);
p = f/pow(1+(i/100),n);
printf("需要投入的本金为：%0.2lf\n",p);
}

main()
{
int num=0;
printf("\t\t\t|***********复利计算器************|\n");
printf("\t\t\t|---------------------------------|\n");
printf("\t\t\t| 1.单利 |\n");
printf("\t\t\t| 2.复利 |\n");
printf("\t\t\t| 3.计算本金 |\n");
printf("\t\t\t|---------------------------------|\n");
printf("请输入选择：");
scanf("%d",&num);
switch(num)
{
case 1:
Danli();
break;
case 2:
Fuli();
break;
case 3:
Need();
default:
return 0;
}
}