A <span>stagflation is my answer.</span>
Solution :
#include
#include
#include
//Converts
to binary string.
* hexadecimal
Binary(char* hexdec)
{
long
= 0;
char *string =
(sizeof(char) * 9);
while (hexdec[i]) {
//Simply assign binary string for each hex char.
switch (hexdec[i]) {
strcat(string, "0000");
break;
strcat(string, "0001");
break;
strcat(string, "0010");
break;
strcat(string, "0011");
break;
strcat(string, "0100");
break;
strcat(string, "0101");
break;
strcat(string, "0110");
break;
strcat(string, "0111");
break;
strcat(string, "1000");
break;
strcat(string, "1001");
break;
case 'A':
case 'a':
strcat(string, "1010");
break;
case 'B':
case 'b':
strcat(string, "1011");
break;
case 'C':
case 'c':
strcat(string, "1100");
break;
case 'D':
case 'd':
strcat(string, "1101");
break;
case 'E':
case 'e':
strcat(string, "1110");
break;
case 'F':
case 'f':
strcat(string, "1111");
break;
default:
printf("\nInvalid hexadecimal digit %c",
hexdec[i]);
string="-1" ;
}
i++;
}
return string;
}
int main()
{ //Take 2 strings
char *str1 =hexadecimalToBinary("FA") ;
char *str2 =hexadecimalToBinary("12") ;
//Input 2 numbers p and n.
int p,n;
scanf("%d",&p);
scanf("%d",&n);
//keep j as length of str2
int j=strlen(str2),i;
//Now replace n digits after p of str1
for(i=0;i<n;i++){
str1[p+i]=str2[j-1-i];
}
//Now, i have used c library strtol
long ans = strtol(str1, NULL, 2);
//print result.
printf("%lx",ans);
return 0;
}
Answer:
"Come help me, please" "Can you help me here?"
Explanation:
Some form of question or help, i think...
Answer:The French and Dutch settlements differed from the Spanish colonies in that they were created mainly to trade and develop industries, while the Spanish were primarily concerned with gold and silver excavation, and then later with sugar exportation.
Explanation:The French and Dutch settlements differed from the Spanish colonies in that they were created mainly to trade and develop industries, while the Spanish were primarily concerned with gold and silver excavation, and then later with sugar exportation.