#include #include #include #include int main(void) { int d = 300000, p = 6000; double r = 0.01, m; m = log10(p / (p - d * r)) / log10(1 + r); printf("贷款%d需要%.1lf月能还完", d, m); return 0; }

#include <stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
int main(void)
{
    int d = 300000, p = 6000;
    double r = 0.01, m;
    m = log10(p / (p - d * r)) / log10(1 + r);
    printf("贷款%d需要%.1lf月能还完", d, m);
    return 0;
}

1. #include<stdio.h>
2. #include <math.h>
3.  
4. int main()
5. {
6. float d = 300000, p = 6000, r = 0.01, m;
7. m = log10(p / (p - d*r)) / log10(1 + r);
8. printf("m = %3.1f\n", m);
9. return 0;
10. }

#include<stdio.h>

#include<math.h>

int main(){

int d=300000,p=6000;

float r=0.01,m;

m=(log10(p)-log10(p-d*r))/log10(1+r);

printf("%.1f",m);

return 0;

}

int main() {
    double d = 300000;
    double p = 6000;
    double r = 0.01;
    double m = 0;
    m = log(p / (p - d * r)) / log(1 + r);
    printf("要%.1lf个月后才能还清贷款", m);
}

#include<stdio.h>
#include<math.h>
int main()
{
    int d=300000;
    int p=6000;
    float r=0.01;
    float m=log(p/(p-d*r))/log(1+r);
    printf("%.1f\n",m);
}