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Source code |
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#include <stdio.h>
#define k 0.1693;
int main (){
int m; /*Variablendeklarationen*/
float x, y, z;
y=0;
x=1000000;
for(m=0; m<10; m++) /*1. Schleife multipliziert 10x*/
{
y=y+x*k;
}
z=y; /*z als Kopie für spätere Ergebnissausgabe*/
do /*2. Schleife dividiert 10x*/
{
y=y/k;
m=m-1;
}
while(m<0);
printf("Ausgangswert: %.8f\n", z);
printf("Endwert: %.8f\n\n", y);
getchar();
return 0;
}
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Source code |
1 |
float expf(float x); |
Quoted
NAME
exp, expf, expl - base-e exponential function
SYNOPSIS
#include <math.h>
double exp(double x);
float expf(float x);
long double expl(long double x);
DESCRIPTION
The exp() function returns the value of e (the base of natural loga-
rithms) raised to the power of x.
Quoted
#include <iostream>
#include <stdlib.h>
using namespace std;
int main(int argc, char *argv[])
{
int n; /*Variablendeklarationen*/
float x=1000000, k = 0.1693;
for(n=0; n<10; n++) // 10 mal teilen
x=x*k;
for(n=0; n<10; n++) // 10 mal multiplizieren
x=x/k;
printf("%.8f\n", x);
system("PAUSE");
return 0;
}
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Source code |
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 |
#include <stdio.h>
#define k 0.1693;
int main (){
int m; /*Variablendeklarationen*/
float x, y, z;
x=1000000;
y=x;
for(m=0; m<10; m++) /*1. Schleife multipliziert 10x*/
{
y*=k;
}
for(m=0; m<10; m++) /*2. Schleife dividiert 10x*/
{
y/=k;
}
printf("Ausgangswert: %.8f\n", x);
printf("Endwert: %.8f\n\n", y);
getchar();
return 0;
}
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Quoted from "Draco"
scho gut, ich habs gerafft![]()
Dafür ist mir der letzte Aufgabenteil völlig suspekt. Hat da jemand ne Idee ?
Zur Aufgabenstellung (Bild)
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