Smooth solution for 2nd euler problem

euler2.c

@@ -0,0 +1,15 @@
+#include <stdio.h>
+
+int main( int argc, char *argv[] ) {
+    int fib_even=2,fib_prev=1,fib_tmp;
+    int sum=0;
+
+    while(fib_even < 4e6) {
+        sum += fib_even;
+        fib_tmp = fib_prev + fib_even;
+        fib_prev = fib_even + fib_tmp;
+        fib_even = fib_prev + fib_tmp;
+    }
+    printf("%d\n",sum);
+    return 0;
+}