UVa problem 10783: sample algorithm

For UVa problem 10783 (Odd Sum), one just has to sum up all the odd numbers between two numbers, a and b. Things can get slow if for each number between a and b, one checked if that number were odd or even, like in the following code:

for (i = a, sum = 0; i <= b; i += 1) {
__j = i mod 2;
__if (j == 1) {
____sum += i;
__}
}

To hasten the computation, we can start at an odd number in the for-loop and add 2 to the iterator index for each iteration.

First we check if a is odd or even. If it is even, we increment it to make it odd. Then we go to the for-loop:

for (j = a, sum = 0; j <= b; j += 2) {
__sum += j;
}

UVa problem 10783: sample code

001 #include "stdio.h"
002 
003 int main() {
004   char line[100];
005 
006   int   num_cases
007       , i
008       , j
009       , k
010       , a
011       , b
012       , sum
013       ;
014 
015   fgets(line, 100, stdin);
016   sscanf(line, "%d", &num_cases);
017 
018   for (i = 1; i <= num_cases; i += 1) {
019     fgets(line, 100, stdin);
020     sscanf(line, "%d", &a);
021 
022     fgets(line, 100, stdin);
023     sscanf(line, "%d", &b);
024 
025     k = a & 1;
026     if (k == 0) {
027       a += 1;
028     }
029 
030     for (j = a, sum = 0; j <= b; j += 2) {
031       sum += j;
032     }
033 
034     printf("Case %d: %d\n", i, sum);
035   }
036 
037   return 0;
038 }