M=(㏒P-㏒(P-D*R))/ ㏒( 1+R)
M is the number of months required to pay off the loan. Suppose today D=324 500 yuan, P=3245 Yuan, and R=0.8%. The program asks about the number of months m to repay the loan and how much to pay in total.
# include & ltmath.h & gt
# include & ltstdio.h & gt
Double m (double p, double d, double r)
{
Double a, b, c;
a = log(P)；
b = log(P-D * R)；
c = log( 1+R)；
Return (a-b)/c;
}
int main(void)
{
Double d = 324500, p = 3245, r = .008f
Long month = 0;
month = (long)M(p，d，r)；
printf(" months = % d \ n payment = % d \ n "，month，month *(long)p)；
Returns 0;
}
/* Operation result:
Month = 20 1
Payment = 652245
*/
Write a program to output English letters C, H, I, N and A one by one. Then output in reverse order, that is, a, n, I, h, c.
# include & ltstdio.h & gt
int main(void)
{
Char s[6]= "China"; int I = 0；
for(I = 0； I<5; i++) printf("%c "，s[I])；
printf(" \ n ")；
for(I = 4； I>- 1; i - )printf("%c "，s[I])；
printf(" \ n ")；
Returns 0;
}
3. Enter the three-side lengths A, B and C of the triangle, and write a program to find the area of the triangle. Known triangle area formula is:
Area=sprt(s(s-a)(s-b)(s-c)), where s=(a+b+c)/2.
# include & ltmath.h & gt
# include & ltstdio.h & gt
Double area (double A, double B, double C)
{
Double s = 0；;
s =(a+b+c)/2.0f；
s = s *(s-a)*(s-b)*(s-c)；
Return to sqrt
}
int main(void)
{
Double a, b, c;
Scanf("%f %f %f ",& i, & ampb & amp；; c)；
printf("area = %f "，area(a，b，c))；
Returns 0;
}
4. Write a program to find ax? The root of the equation +bx+c=0. A, b, c input with the keyboard, b? -4ac & gt； 0。
# include & ltmath.h & gt
# include & ltstdio.h & gt
Double area (double A, double B, double C)
{
Double s = 0；;
s =(a+b+c)/2.0f；
s = s *(s-a)*(s-b)*(s-c)；
Return to sqrt
}
int main(void)
{
int a，b，c，d； Double e, x, y;
Scanf("%d %d %d ",& i, & ampb & amp；; c)；
d = b * b-4 * a * c；
if(d & lt； 0) {
Printf ("There is no real root. \ n ")；
Returns 0;
}
if(d == 0) {
e =-2 * a；
e =(double)b/e；
printf("X 1 = X2 = %f "，e)；
Returns 0;
}
e = d；
e = sqrt(e)；
x =(-(double)b+e)/(double)(2 * a)；
y =(-(double)b-e)/(double)(2 * a)；
printf("X 1 = %f，X2 = %f\n "，x，y)；
Returns 0;
}