# Google Code Jam 2016 Round 1B B. Close Match

,

The logic is simple, but the code is long. This is golfable?

## problem

$10$進$N$桁の固定精度整数$A,B$が与えられる。ただしそれらは$10$進表記されており、かつそのいくらかの桁が?で置き換えられている。 ?の桁を全て適当な数字で置き換える。 置き換えた結果$A’,B’$に関し、3つ組$(|A’ - B’|, A’, B’)$が辞書順最小になるように置き換え、$A’,B’$を出力せよ。

## solution

Decide from the more significant digits, pairwise. $O(N^2)$.

Pairwise and recursively see the digits and fill ?s from the left. If the order of the results $A’, B’$ are already decided, you must fill the ?s with 0 or 9 appropriately. Else, it is not decided yet, and there are ?s, you must try two (or three) ways: fill as both digits are same, or one is greater than another by just $1$.

## implementation

#include <iostream>
#include <string>
#include <tuple>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
#define repeat_from(i,m,n) for (int i = (m); (i) < (n); ++(i))
typedef long long ll;
template <class T> bool setmin(T & l, T const & r) { if (not (r < l)) return false; l = r; return true; }
using namespace std;
const ll inf = 1e18+9;
tuple<ll,ll,ll> dfs(int i, string const & s, string const & t, ll a, ll b) {
int n = s.length();
if (i == n) {
return make_tuple(llabs(a - b), a, b);
} else if (a > b) {
ll na = a * 10 + (s[i] == '?' ? 0 : s[i] - '0');
ll nb = b * 10 + (t[i] == '?' ? 9 : t[i] - '0');
return dfs(i+1, s, t, na, nb);
} else if (a < b) {
ll na = a * 10 + (s[i] == '?' ? 9 : s[i] - '0');
ll nb = b * 10 + (t[i] == '?' ? 0 : t[i] - '0');
return dfs(i+1, s, t, na, nb);
} else {
auto ans = make_tuple(inf, inf, inf);
if (s[i] == '?' and t[i] == '?') {
setmin(ans, dfs(i+1, s, t, a * 10 + 0, b * 10 + 0));
setmin(ans, dfs(i+1, s, t, a * 10 + 0, b * 10 + 1));
setmin(ans, dfs(i+1, s, t, a * 10 + 1, b * 10 + 0));
} else if (s[i] == '?') {
ll nb = b * 10 + (t[i] - '0');
repeat_from (j,-1,1+1) if (isdigit(t[i] + j)) {
ll na = a * 10 + (t[i] + j - '0');
setmin(ans, dfs(i+1, s, t, na, nb));
}
} else if (t[i] == '?') {
ll na = a * 10 + (s[i] - '0');
repeat_from (j,-1,1+1) if (isdigit(s[i] + j)) {
ll nb = b * 10 + (s[i] + j - '0');
setmin(ans, dfs(i+1, s, t, na, nb));
}
} else {
ll na = a * 10 + (s[i] - '0');
ll nb = b * 10 + (t[i] - '0');
setmin(ans, dfs(i+1, s, t, na, nb));
}
return ans;
}
}
string zfill(string const & s, int n) {
string t(n - s.length(), '0');
return t + s;
}
void solve() {
string s, t; cin >> s >> t;
int n = s.length();
ll diff, a, b; tie(diff, a, b) = dfs(0, s, t, 0, 0);
cout << zfill(to_string(a), n) << ' ' << zfill(to_string(b), n) << endl;
}
int main() {
int t; cin >> t;
repeat (i,t) {
cout << "Case #" << i+1 << ": ";
solve();
}
return 0;
}