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记念一下~~~这题很简单考基础的#include <stdio.h>
int jie(int n )
{ int sum = 1 , i ;
for( i = 1 ; i <= n ; i++ )
sum *= i ;
return sum ;
}
int main()
{
int n = 9 , i = 0 ;
double sum = 1 ;
printf("n e\n") ;
printf("- -----------\n") ;
printf("0 1\n1 2\n2 2.5\n3 2.666666667\n4 2.708333333\n") ;
for( i = 1 ; i <= 4 ; i++ )
{ sum += 1.0 / jie(i) ;
}
for( i = 5 ; i <= n ; i++ )
{ sum += 1.0 / jie(i) ;
printf("%d %.9lf\n",i , sum ) ;
}
return 0 ;
}
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