给出 T，n， A， K，a， b， m， P.
T 组样例.
求
由于
所以
如果用快速幂求和的话会 TLE.
因为
所以只需要求 sum1[] 和 sum2[].
sum1[i]：
sum2[i]：
详见代码.
#include #include #include #include #include #include #include #include #include #include #include #include #include #define eps 1e-6 #define PI acos(-1.0) #define ll long long int using namespace std; const int M = (int)1e5; ll n, A, K, a, b, m, p; ll sum1[M + 5]; ll sum2[M + 5]; void init() { sum1[0] = sum2[0] = 1; for(int i = 1; i <= M; ++i) sum1[i] = sum1[i - 1] * A % p; sum2[1] = sum1[M]; for(int i = 1; i <= M / 10; ++i) sum2[i] = sum2[i - 1] * sum2[1] % p; } ll f() { ll fx = K; ll sum = 0; for(ll i = 1; i <= n; ++i) { sum = (sum1[fx % M] * sum2[fx / M] % p + sum) % p; fx = (a * fx + b) % m; } return sum % p; } int main() { int T; scanf("%d", &T); for(int ca = 1; ca <= T; ++ca) { cin >> n >> A >> K >> a >> b >> m >> p; init(); ll ans = f(); printf("Case #%d: %lld\n", ca, ans); } return 0; }
![A^{f(x)} = sum1[i \;mod\;M] \times sum2[i\;/\;M]](http://img.e-com-net.com/image/info8/aa1caf341d704dd9936df16591a24009.gif)
