[案例74]计算s=1k+2k+3k++N /*案例代码文件名:AL74C* /*功能:函数的嵌套调用* #define K 4 #define n 5 long fi(int n, int k) /*计算n的k次方* i long power=n int 1 for(i-l; i<k; i++)power *=n; return power
[案例7.4] 计算s=1 k+2 k+3 k+……+N k /*案例代码文件名：AL7_4.C*/ /*功能：函数的嵌套调用*/ #define K 4 #define N 5 long f1(int n,int k) /*计算n的k次方*/ { long power=n; int i; for(i=1;i<k;i++) power *= n; return power; }
long f2(int n, int k) /*计算1到n的k次方之累加和* f long sI t for(i-l; K<=n; i++)sum+=fl(,k); return sum main( i printf("Sum of %d powers of integers from 1 to %d="K, N) printf("%odn",f2(N, K )) [程序演示]
long f2(int n,int k) /*计算1到n的k次方之累加和*/ { long sum=0; int i; for(i=1;i<=n;i++) sum += f1(i, k); return sum; } main() { printf("Sum of %d powers of integers from 1 to %d = ",K,N); printf("%d\n",f2(N,K)); getch(); } [程序演示]