void power(float P,float R,float n,float A);
void main()
{
float A,P,R,n;
clrscr();
printf("Enter principal amount:");
scanf("%f",&A);
printf("Enter rate of interest:");
scanf("%f",&R);
printf("Enter number of years:");
scanf("%f",&n);
power(A,R,P,n);
getch();
}
void power(float A,float R,float P,float n)
{
int i;
for(i=1;i<=n;i++)
{
A=1+(R*0.01);
A=A*i;
A=A*P;
printf("For year %d, C.I =%f\n",i,A);
}
}
I put
Principal = 2000
Rate of interest=3
Years=3
我得到的复利值为负数。错误在于A=A*P;必须计算每年的复利,而不使用幂函数
最佳答案
#include<stdio.h>
void power(float P,float R,float n,float A);
void main()
{
float A,P,R,n;
// clrscr();
printf("Enter principal amount:");
scanf("%f",&A);
printf("Enter rate of interest:");
scanf("%f",&R);
printf("Enter number of years:");
scanf("%f",&n);
power(A,R,P,n);
// getch();
}
void power(float A,float R,float P,float n)
{
int i;
float intr=0;
for(i=1;i<=n;i++)
{
intr=0;
intr=A*(R/100);
A=A+intr;
// A=A*P;
printf("For year %d, C.I =%f\n",i,intr);
}
}
关于计算复利并显示每年的利息,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38804462/