1.for循环实现
1 #include <stdio.h>
2 #include <time.h>
3
4 clock_t start, stop;
5 double duration;
6
7 void printfN();
8
9 int main(){
10 int N;
11 printf("Pleade enter a number: ");
12 scanf("%d", &N);
13 start = clock();
14 printfN(N);
15 stop = clock();
16 duration = ((double)(stop - start))/CLK_TCK;
17 printf("duration = %f\n",duration);
18 return 0;
19 }
20
21 void printfN(int N){
22 int i;
23 for(i=0; i<=N; i++){
24 printf("%d\n",i);
25 }
26 }
2.递归实现（有问题，思考——算法的执行效率与输入规模的关系）
1 #include <stdio.h>
2 #include <time.h>
3
4 void printfN();
5
6 clock_t start, stop;
7 double duration;
8
9 int main(){
10 int N;
11 printf("Please enter a number:");
12 scanf("%d", &N);
13 start = clock();
14 printfN(N);/*12000程序无法运行*/
15 stop = clock();
16 duration = ((double)(stop - start))/CLK_TCK;
17 printf("duration = %f\n",duration);
18 return 0;
19 }
20
21 void printfN(N){
22 if(N){
23 printfN(N-1);
24 }
25 printf("%d\n",N);
26 }