复利计算器2

#include<stdio.h>
#include<math.h>
void tip()
{
printf("复利计算软件\n");
printf(" 复利计算公式\n");
printf("\tF=P*pow((1+i/m),N*m)\n");
printf(" 本金计算公式\n");
printf("\tP=F/pow((1+i/m),N*m)\n");
printf(" 单利计算公式\n");
printf("\tL=P*N*i\n\tH=L+P\n");
printf("\tF:复利终值\n");
printf("\tP:本金\n");
printf("\ti:利率\n");
printf("\tH:本利和\n");
printf("\tN：利率获取时间的整数倍\n");
printf("\tm:年复利次数(一年当中的滚利次数)\n");
}
void menu()
{
printf("\t\t------------- CHOOSE ------------\n");
printf("\t\t 1:计算本金 \n");
printf("\t\t 2:计算年复利终值 \n");
printf("\t\t 3:单利计算 \n");
printf("\t\t 4:计算时间 \n");
printf("\t\t 5:计算利率 \n");
printf("\t\t 6:年金终值 \n");
printf("\t\t 0:结束 \n");
printf("\t\t-----------------------------------\n");
printf("请输入你要选择的计算方式(0~3):\n");

}

void benjin()
{
int N,m;
double i,F,P;
printf("复利终值:");
scanf("%lf",&F);
printf("年利率:");
scanf("%lf",&i);
printf("存入年限:");
scanf("%d",&N);
printf("年复利次数:");
scanf("%d",&m);
P=F/pow((1+i/m),N*m);
printf("年复利终值为%.4lf需要本金为:%.4lf\n",F,P);
}
void fuli()
{
int N,m;
double i,F,P;
printf("存入本金:");
scanf("%lf",&P);
printf("年利率:");
scanf("%lf",&i);
printf("存入年限:");
scanf("%d",&N);
printf("年复利次数:");
scanf("%d",&m);
F=P*pow((1+i/m),N*m);
printf("复利终值:%.4lf\n",F);
}
void danli()
{
int N;
double i,H,P,L;
printf("存入本金:");
scanf("%lf",&P);
printf("年利率:");
scanf("%lf",&i);
printf("存入年限:");
scanf("%d",&N);
L=P*N*i;
H=L+P;
printf("本息和为:%.4lf\n",H);
}

void shijian()
{
int N,m;
double i,F,P;
printf("存入本金:");
scanf("%lf",&P);
printf("年利率:");
scanf("%lf",&i);
printf("复利终值:");
scanf("%lf",&F);
printf("年复利次数:");
scanf("%d",&m);
N=(int)(log(F/P)/log(1+i/m))/m;
printf("时间:%d\n",N);
}
void lilu()
{
int N,m;
double i,F,P;
printf("存入本金:");
scanf("%lf",&P);
printf("存入年限:");
scanf("%d",&N);
printf("复利终值:");
scanf("%lf",&F);
printf("年复利次数:");
scanf("%d",&m);
i=m*pow(F/P,1.0/N*m)-m;
printf("利率:%.4lf\n",i);
}
void nianjinzhongzhi()//计算年金终值
{
int N,n;
double i,F,P;
printf("存入本金:");
scanf("%lf",&P);
printf("存入年限:");
scanf("%d",&N);
printf("年利率:");
scanf("%lf",&i);
printf("\t\t1:按年投资\n\t\t2:按月投资\n");
A:printf("请选择你要的功能<1|2>:");
scanf("%d",&n);
if(n==1)
{
F=P*(pow(1+i,N)-1)/i;

}
else if(n==2)
{
F=N*12*(P*(i/12)+P);
}
else
{
printf("输入有误!请重新输入\n");
goto A;
}
printf("%d年后的总产值:%.4lf\n",N,F);

}
void main()
{
int n;
while(1)
{
tip();
menu();
scanf("%d",&n);
if(n==0)
break;
switch(n)
{
case 1:
benjin();break;
case 2:
fuli();break;
case 3:
danli();break;
case 4:
shijian();break;
case 5:
lilu();break;
case 6:
nianjinzhongzhi();break;
case 0:n=0;break;
}

}
}