CodeForces - 964C（等比数列求和+逆元）

链接：CodeForces - 964C

题意：求 ${n\sum i=0sian-ibi\sum i=0nsian-ibi by 109+9109+9\; 值。s[i]为+1或-1。}^{}$

${题解：可证：每k项的和相差(b/a)^k。等比数列求和加逆元。}^{}$

${逆元不一定要最后一步用的，而是每步都可以用。}^{}$

#include 
using namespace std;

const double EPS = 1e-8;
const int mod = 1e9 + 9;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 10;
long long n, a, b, k;
char s[maxn];

long long pow_mod(long long x, long long n)
{
    long long res = 1;
    while(n){
        if(n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}

void Exgcd(long long a, long long b, long long& x, long long& y)
{
    if(!b){
        y = 0; x = 1;
        return ;
    }
    Exgcd(b, a % b, y, x);
    y -= a / b * x;
}

long long NY(long long a, long long b)
{
    long long x, y;
    Exgcd(a, b, x, y);
    return (x % mod + mod) % mod;
}

int main()
{
    scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
    scanf("%s", s);

    long long ans = 0;
    for(int i = 0; i < k; i++){
        long long tmp = pow_mod(a, n - i) * pow_mod(b, i) % mod;
        if(s[i] == '+') ans += tmp;
        else ans -= tmp;
        ans = (ans % mod + mod) % mod;
    }

    long long q = pow_mod(b, k) * NY(pow_mod(a, k), mod) % mod;
    long long m = (n + 1) / k;

    if(q == 1) ans = ans * m % mod;
    else ans = ans * (pow_mod(q, m) - 1) % mod * NY(q - 1, mod) % mod;

    ans = (ans % mod + mod) % mod;

    printf("%lld\n", ans);;

    return 0;
}