Given a string, computes the number of values in the string and sums them. It also contains the minus sign-if the minus sign is a numeric value, then it is a negative number, if followed by not numbers, it does not mean what input: A string output: Numeric values and example input: 312ab-2---9--a
Output: 3 301
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
intStrtoint (Char* str)
{
intSign,sum,k,i,len,num;
Sign =1;
sum = i =0;
len = strlen (str);
if(str[i]=='-')
{
sign =-sign;
i++;
}
if(str[i]=='+')
{
i++;
}
num =0;
for(; i<len;i++)
{
//sum + = Sum*int (Pow (10,num)) + (str[i]-' 0 ');
sum = sum*Ten+ (str[i]-'0');
//num++;
}
returnSum*sign;
}
intMain ()
{
intI,len,j;
Charstr[1000001],tep[1000001],c;
Doubleb =0.9999;
intsum =0;
Gets (str);
len = strlen (str);
//i = atoi (str);
//B = atof (str);
//printf ("%lf\n", b);
//sprintf (str, "%.3lf hello", b);
//puts (str);
j =0;
for(i=0; i<len;i++)
{
c = Str[i];
if(str[i]=='-')
{
if(j>0)
{
Sum+=strtoint (TEP);//atoi (TEP);
memset (TEP,0,sizeof(TEP));
}
j=0;
if(str[i+1]>='0'&&str[i+1]<='9')
{
tep[0]='-';
j + +;
}
}
Else
{
if(str[i]>='0'&&str[i]<='9')
{
TEP[J] = Str[i];
j + +;
//if (i==len-1)
}
Else
{
tep[j]=' /';
if(j>0)
{
Sum+=strtoint (TEP);//atoi (TEP);
memset (TEP,0,sizeof(TEP));
j=0;
}
}
}
}
if(j>0)
{
Sum+=strtoint (TEP);
}
printf"%d\n", sum);
return0;
//312df-2ff--9
}
Sum the numbers in a string
