Determine the number of prime numbers between-and output all prime numbers.

Question: determine the number of prime numbers between-and output all prime numbers.

Program Analysis: first understand what is a prime number, only the number divisible by 1 and itself, traverse the number between 101 and in a loop, and then use ~ For example, if the number is 200, we divide the number from 2 to 2 ~ The number between 112, as long as the integer here is not equal to 0, you can judge that this number is a prime number;

# Include <stdio. h> int main () {int I, j; int COUNT = 0; for (I = 101; I <= 200; I ++) {for (j = 2; j <I; j ++) {// If J can be rounded up by I in the out-of-loop if (I % J = 0) break;} // determine whether the loop exists in advance, if j <I indicates 2 ~ Between J, I has an integer. If (j> = I) {count ++; printf ("% d", I); // line feed, count with count, if (count % 5 = 0) printf ("\ n") ;}} return 0 ;}

Running result:

The other is to use a number to remove 2 to SQRT (This number). If it can be divisible, it indicates that this number is not a prime number, and vice versa.

#include <stdio.h>#include <math.h>int main(){    int i,j,k,leap=1;    int count=0;        for (i=101; i<=200; i++)     {        k=sqrt(i+1);        for (j=2; j<=k; j++)         {            if (i%j==0)             {                leap=0;                break;            }        }                if (leap)         {            count++;            printf("%d ",i);            if (count % 5 == 0)             printf("\n");        }                leap=1;    }        return 0;}

Running result: