单复利计算加强版

Posted 2020-06-21 50黄柏堂

#include<stdio.h>
#include<math.h>

void fuli(){
int n,i;
double ll,sum,bj;
system("cls");
printf("\t\t\t欢迎进入单复利计算:\n");
printf("请输入本金:");
scanf("%lf",&bj);
printf("输入所要存储年:");
scanf("%d",&n);
printf("输入利率:");
scanf("%lf",&ll);
printf("输入复利次数:");
scanf("%d",&i);
printf("正在输出:");
ll=pow((1+ll/i),i)-1;
sum=bj*(pow(1+ll,n));
printf("所得%lf",sum);
printf("\n");
}

void danli(){
int n;
double ll,sum,bj;
system("cls");
printf("\t\t\t欢迎进入单复利计算:\n");
printf("请输入本金:");
scanf("%lf",&bj);
printf("请输入存储年:");
scanf("%d",&n);
printf("输入利率:");
scanf("%lf",&ll);
printf("正在输出:");
sum=bj+bj*n*ll;
printf("所得%lf",sum);
printf("\n");
}

void benjin(){
int n,i;
double ll,sum,bj;
system("cls");
printf("\t\t\t欢迎进入单复利计算:\n");
printf("请输入你所需要的总值:");
scanf("%lf",&sum);
printf("请输入你所需要的存储年:");
scanf("%d",&n);
printf("请输入利率:");
scanf("%lf",&ll);
printf("请输入复利次数:");
scanf("%d",&i);
ll=pow((1+ll/i),i)-1;
bj=sum/pow(1+ll,n);
printf("正在输出:");
printf("所需要本金%lf",bj);
printf("\n");
}

main(){
int a;
printf("\t\t\t欢迎进入单复利计算:\n");
printf("\t1.单利计算");
printf("\n\t2.复利计算");
printf("\n\t3.求本金");
printf("\n请输入你的选项:");
scanf("%d",&a);
switch (a){
case 1:
danli();
break;
case 2:
fuli();
break;
case 3:
benjin();
break;
default:
return 0;
}
}