複利計算3

#include<stdio.h>

#include<math.h>

void fuli()

{

int day;

double total = 1.0f;

float money;

float base;

double total1;

printf("------------------------------------------------------------\n");

printf("每年的年利率比為:");

scanf("%f",&base);

printf("輸入你存入的年數:");

scanf("%d",&day);

total=total*pow(1+base,day);

printf("-------------------------------------------------------------\n");

printf("每年的年利率%f,連續%d年,總受益的利率:%lf\n",base,day,total);

printf("輸入本金金額:");

scanf("%f",&money);

total1=money*total;

printf("你将可以得到:%lf\n",total1);

}

void danli()

total=base*day;

total1=money+money*total;

void fanxiang()

float money;

float base;

double total = 1.0f;

int day;

int x;

double total1;

printf("請輸入你需要得到的金額:\n");

scanf("%f",&money);

printf("請輸入平均年回報率:\n");

scanf("%f",&base);

printf("請輸入存入的年數\n");

scanf("%d",&day);

printf("若是複利選擇1，單利選擇2\n");

scanf("%d",&x);

if(x==1)

{

total=total*pow(1+base,day);

total1=money/total;

printf("需要的本金為:%f\n",total1);

}

else

total=base*day;

total1=money/(1+total);

void gupiao()

double base=0.1;

float total;

double day;

printf("請輸入你的本金:\n");

scanf("%f",&total);

printf("此股票平均年報酬率為(百分之十)!\n");

day=(money-total)/(total*base);

printf("需要的年數為:%.3lf年\n",day);

void ligunli()

int i;

double total2;

double total3=0;

printf("若是月利選擇1，年利選擇2\n");

printf("請輸入你的本金:\n");

scanf("%lf",&total1);

printf("請輸入平均月回報率:\n");

printf("請輸入存入的月數\n");

for(i=1;i<=day;i++)

total2=total1+(total1*base);

total1=total2;

total3=total1+total3;

printf("最後可以得到%.3lf元\n",total3);

{

for(i=1;i<=day;i++)

void touzi()

printf("請輸入存入的年數:\n");

scanf("%lf",&day);

base=(money-total)/(day*total);

printf("投資工具報酬率需要有%.3lf才夠\n",base);

void main()

int num=4;

printf(" \n");

printf(" -------------------------------------\n");

printf(" 歡迎來到廣州商業銀行\n");

printf(" 顧客你可以根據1~3選擇\n");

printf(" 1.複利計算\n");

printf(" 2.單利計算\n");

printf(" 3.送你養老金計算方法\n");

printf(" 4.股票賺錢喲\n");

printf(" 5.投資工具找的好才賺錢喲\n");

printf(" 6.本利與本金的存款計算\n");

printf("請根據數字選擇你需要的功能(1~6)\n");

scanf("%d",&num);

switch(num)

{

case 1:

fuli();

break;

case 2:

danli();

case 3:

fanxiang();

case 4:

gupiao();

case 5:

touzi();

case 6:

ligunli();

}