静态局部变量:1~5的阶乘 计算s=1k+2k+3k+……+Nk
(2010-04-28 10:20:10)
标签:
杂谈 |
#include
<stdio.h>
int fac(int n)
{
static int f=1;
f=f*n;
return f;
}
void main()
{
int i;
for(i=1; i<6;
i++)
printf("%d!=%d\n",i,fac(i));
}
==============================================
//应用举例:计算s=1k+2k+3k+……+Nk
#include <stdio.h>
#define K 4
#define N 5
long f1(int n,int k);
long f2(int n,int k);
void main()
{
printf("Sum of %d powers of integers from 1 to %d
= ",K,N);
printf("%d\n",f2(N,K));
}
long f1(int n,int
k)
{
long power=n;
int i;
for(i=1;i<k;i++)
power *= n;
return power;
}
long f2(int n,int
k)
{
long sum=0;
int i;
for(i=1;i<=n;i++)
sum += f1(i, k);
return sum;
}
#include <stdio.h>
#define K 4
#define N 5
long
long
void main()
{
}
long
{
}
long
{
}