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@Mistat
Created July 19, 2010 11:17
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#include<stdio.h>
/**
* 自然対数の底eは無理数で、その値は下記の無限級数で表される。
* 小数点3桁までの精度でeの近似値を計算するプログラムを作成
* e=1+1/1!+1/2!+1/3!・・・・・・+1/n!・・・・・
*/
void main()
{
double sum,sum0;
int i,k,kai;
sum0=0;
sum=1;
k=0;
while(sum-sum0>0.001)
{
k++;
kai=1;
for(i=1;i<k;i++) kai=kai*i;
sum0=sum;
sum+=1.0/kai;
}
printf("e=%f\n",sum);
}
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