CF963A Alternating Sum

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const ll MOD = 1e9 + 9;
 5 ll mypow(ll x, ll n)
 6 {
 7     ll ans = 1;
 8     while (n)
 9     {
10         if (n & 1) ans = ans * x % MOD;
11         x = x * x % MOD;
12         n >>= 1;
13     }
14     return ans;
15 }
16 ll inv(ll x)
17 {
18     return mypow(x, MOD - 2);
19 }
20 ll cal(ll x, ll a, ll b, ll n, ll k)
21 {
22     ll q = mypow(b, k) * inv(mypow(a, k)) % MOD;
23     if (q == 1) return x * n % MOD;
24     return x * (mypow(q, n) - 1 + MOD) % MOD * inv(q - 1) % MOD;
25 }
26 int main()
27 {
28     ll n, a, b, k, x, q, l;
29     string s;
30     while (cin >> n >> a >> b >> k >> s)
31     {
32         ll ans = 0, l = (n + 1) / k;
33         for (int i = 0; i < k; i++)
34         {
35             x = mypow(a, n - i) * mypow(b, i) % MOD;
36             ll tmp = (s[i] == ‘+‘ ? 1 : -1) * cal(x, a, b, l, k);
37             ans = (ans + tmp) % MOD;
38             ans = (ans + MOD) % MOD;
39         }
40         cout << ans << endl;
41     }
42     return 0;
43 }