Programming Zhu Ji Nanxiong (cont.) Reading notes-(preface + chapter I performance monitoring tools)

"ACM Newsletter"

One chapter at a time, read carefully

ANSI American National Standards Institute American Standards Society

1.1 Calculating prime numbers

#include <stdio.h>intPrimeintN) {    inti;  for(i =2; i<n;i++) {999if(n%i==0) 78022return 0; 831return 1; 168}}main () {intI, N; N= +; 1 for(i=2; i<=n;i++) 1 {if(Prime (i)) 999 {printf ("%d", i); 168} }}

Prime concept:

1. In all integers larger than 1, except 1 and itself, there are no more approximations, such integers are called prime numbers or primes. It can also be said that a prime number is only 1 and it itself is two approximate.

2. The prime number is such an integer, except that it can be expressed as a product of itself and 1, which cannot be represented as either
  The product of any other two integers.

P2 consider that n a possible integer factor that does not exceed ∫n, the program becomes effective    and the number of calls is reduced

#include <stdio.h>#include<math.h>intRootintN) {    return(int) sqrt (float) n); 5456}intPrimeintN) {    inti;  for(i =2; I<root (n); i++) 999 {if(n%i==0) 5288
　　　　 return 0; 831return 1; 168}}main () {intI, N; N= +; 1 for(i=2; i<=n;i++) {1if(Prime (i)) 999 printf ("%d", i); 168}}

Procedure trial Teaching Performance monitoring instructions, SQRT takes up the most time. P3 move the SQRT function outside the For loop,

#include <stdio.h> #include <math.h>int root (int n) {    return (int) sqrt ((float) n); 5456}int Prime (int n) {    int i,bound;    Bound =root (n);    for (i =2;i<bound;i++)                 999    {        if (n%i==0)                          5288 return 0;                 831        return 1;  168    }} Main () {    int i, n;    n=1000;                                  1    for (i=2;i<=n;i++) {                       1    if (prime (i))                             999        printf ("%d", I); 168    }}

P4

By a special test that is divisible by 2, 3, 5. Eliminate the number of 1/2,1/3,1/5, but there is a problem, 2,3,5 in this program does not output

#include <stdio.h>#include<math.h>intRootintN) {    return(int) sqrt (float) n);}intPrimeintN) {    intI,bound; if(n%2==0)        return 0; if(n%3==0)        return 0; if(n%5==0)        return 0; Bound=root (n);  for(i =7; i<=bound;i=i+2)    {    if(n%i==0)return 0; return 1; }}main () {intI,n; N= +;  for(i =2; i<=n;i++){        if(Prime (i)) printf ("%d", i); }}

The correct test is

if (n%2==0)
return (n==2); The same is true for 3 5.

P5 took the time-consuming prescription and replaced it with multiplication.

Programming Zhu Ji Nanxiong (cont.) Reading notes-(preface + first chapter performance monitoring tools)