如果你不知道怎么来的，百度一下‘贷款利息公式’就行了
变形得到(1+R)^M=P/(P-D*R)
两边同时取对数，得到m*log10(1+R)=log10(P/(P-D*R))=log10(P)-log10(P-D*R)
再变形得到m=m=(log10(p)-log10(p-d*r))/log10(1+r)
/** main.c
*
* Created on: 2011-6-8
* Author: icelights
*/
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <math.h>
#define APR10.0747/*<1年（含1年）年利率*/
#define APR20.0756/*1-3年（含3年）年利率*/
#define APR30.0774/*3-5年（含5年）年利率*/
#define APR40.0783/*5年以上年利率*/
#define A_TO_M 1/12 /*月利率 = 年利率 / 12*/
#define RTP 12/*Reimbursement total periods还款总期数 =年限*12*/
#define LENGTH 80
struct LoanInfo
{
/*姓名*/
char name[LENGTH]
/*贷款总额*/
doubleLoanAmount
/*贷款年限*/
doubleLoanYear
/*月付*/
doubleMonthlyPayment
/*总利息*/
doubleTotalInterest
/*还款总额*/
doubleReimbursementAmount
/*年利率*/
doubleapr
struct LoanInfo * next
}
void CalcShow(struct LoanInfo * cur, struct LoanInfo * hd,
struct LoanInfo * prv)
int main(void)
{
int temp
struct LoanInfo * head = NULL
struct LoanInfo * prev, * current
current = (struct LoanInfo *)malloc(sizeof(struct LoanInfo))
if (NULL == head)
{
head = current
}
else
{
prev->next = current
}/*End of if (NULL == head)*/
puts("请输入姓名")
gets(current->name)
fflush(stdin)
puts("请输入贷款数额（单位：万元）")
scanf("%lf", ¤t->LoanAmount)
fflush(stdin)
puts("请输入贷款年限")
scanf("%lf", ¤t->LoanYear)
fflush(stdin)
printf("姓名：%s,贷款年限：%lf, 贷款数额%lf",
current->name, current->LoanYear, current->LoanAmount)
prev = current
puts("请确认Y/N")
temp = getchar()
switch(toupper(temp))
{
case 'Y' : CalcShow(current, head, prev)
break
case 'N' : free(current)
main()
break
default: puts("输入错误")
free(current)
break
}
return 0
}
void CalcShow(struct LoanInfo * cur, struct LoanInfo * hd,
struct LoanInfo * prv)
{
char lcv_temp
if (cur->LoanYear <= 1)
cur->apr = APR1
else if (cur->LoanYear <= 3)
cur->apr = APR2
else if (cur->LoanYear <= 5)
cur->apr = APR3
else
cur->apr = APR4
/*End of if (year <= 1)*/
cur->LoanAmount = 10000 * cur->LoanAmount
cur->ReimbursementAmount = cur->LoanAmount * pow((1 + cur->apr), cur->LoanYear)
cur->MonthlyPayment = cur->ReimbursementAmount / (cur->LoanYear * RTP)
cur->TotalInterest = cur->ReimbursementAmount - cur->LoanAmount
printf("姓名：%s 贷款年限：%.0lf\n"
"贷款数额：%.2lf 每月还款额：%.2lf\n"
"利息合计：%.2lf 还款总额：%.2lf\n",
cur->name, cur->LoanYear, cur->LoanAmount,
cur->MonthlyPayment, cur->TotalInterest, cur->ReimbursementAmount)
puts("是否继续计算Y/N")
lcv_temp = getchar()
switch(toupper(lcv_temp))
{
case 'Y' : free(cur)
main()
break
case 'N' : free(cur)
exit(0)
default: puts("输入错误")
free(cur)
main()
break
}
system("pause")
}
#include <stdio.h>float cal_power(float x, int n)
{
float p=1.0
while(n>0) {
p=p*x
n--
}
return p
}
float cal_money(int loan, float rate, int month)
{
double tmp
tmp=cal_power(1+rate,month)
return loan*rate*tmp/(tmp-1)
}
int main(void)
{
int loan,year,month
float money,rate
printf("Enter loan: ")
scanf("%d",&loan)
printf("Enter rate: ")
scanf("%f",&rate)
for(year=5year<=30year++) {
month=12*year
money=cal_money(loan,rate,month)
printf("money(%d,%d)=%.0f\n",loan,year,money)
}
return 0
}
