[Algorithm] [look] dichotomy Bisection

#include <stdio.h> int main () {
     int A [] = { . 1 , 2 , . 3 , . 4 , . 5 , . 6 , . 7 , . 8 , . 9 , 10 , . 11 , 12 is , 13 is , 14 , 15 };
     int = L the sizeof (a) / . 4 ;         // length int X = 10 ;                 // to find the value of int Bisection ( int X,


    
    
    int * A, int L); 
    Bisection (X, A, L); 
    
    return  0 ; 
} 

int Bisection ( int X, int * A, int L) {
     int address;     // find the location of the number of 
     // consider two end position 
    IF (X == a [L- . 1 ]) { 
        address = L- . 1 ; 
        the printf ( " % D \ n- " , address);
         return address; 
    } the else  IF (a == X [ 0 ]) { 
        address =0;
        printf("%d\n",address);
        return address;
    }
    //如果不在两端 
    address=l/2;
    l=l/2;
    while(true){
        if(x==a[address]){
            printf("%d\n",address);
            break;
        }else if(x>a[address]){
            l=l/2;
            address=address+L; 
        } the else  IF (X < A [address]) { 
            L = L / 2 ; 
            address = Address- L; 
        } 
        IF (L == 0 ) { 
             // If l = 0 can not be found indicating 
            the printf ( " Not found " );
             BREAK ; 
        } 
    }      
    return address; 
}