Java: 92 987 additions
I use binary recursion, that a/b == 2 * a/(2b) + maybe 1. For that divisor and remainder are needed. There would normally be a subtraction a % (2b) - b, but that is resolved by holding the remainder as (rem, remNegative). And 2b = b+b of course.
static int add_used;static int add(int a, int b) { if (a == 0) return b; if (b == 0) return a;++add_used; return a + b;}private static class DivRem { int div; int rem; int remNegative; DivRem(int div, int rem) { this.div = div; this.rem = rem; }}public static int divide(int a, int b) { add_used = 0; return divrem(a, b).div;}public static DivRem divrem(int a, int b) { if (b > a) { return new DivRem(0, a); } DivRem dr = divrem(a, add(b, b)); dr.div = add(dr.div, dr.div); if (dr.rem >= add(b, dr.remNegative)) { dr.div = add(dr.div, 1); //dr.rem = add(dr.rem, -b); dr.remNegative = add(dr.remNegative, b); } return dr;}private static void test(int a, int b) { boolean okay = a/b == divide(a, b); System.out.printf("%d / %d = %d :: %d : #%d %s%n", a, b, a/b, divide(a, b), add_used, okay);}public static void main(String[] args) { //test(2352, 324); int n = 0; for (int a = 1; a <= 200; ++a) { for (int b = 1; b <= a; ++b) { //test(a, b); divide(a, b); n += add_used; } } System.out.println("Additions: "+ n);}