The 2019 ICPC Asia Yinchuan Regional Programming Contest/2019銀川區域賽 D Easy Problem(莫比烏斯容斥+尤拉降冪)
阿新 • • 發佈:2020-11-24
The 2019 ICPC Asia Yinchuan Regional Programming Contest/2019銀川區域賽 D Easy Problem(莫比烏斯容斥+尤拉降冪)
題面:
題意:
給定4個整數\(n,m,d,k\),讓你求\(\sum_{a_1=1}^{m}\sum_{a_2=1}^{m}\dots\sum_{a_n=1}^{m}[gcd(a_1,a_2,\dots,a_n)=d](a_1*a_2*\dots*a_n)^k\)。
思路:
利用莫比烏斯函式性質和乘法分配律得:
\[\sum_{a_1=1}^{m}\sum_{a_2=1}^{m}\dots\sum_{a_n=1}^{m}[gcd(a_1,a_2,\dots,a_n)=d](a_1*a_2*\dots*a_n)^k \\ =d^{kn}\sum_{a_1=1}^{m/d}\sum_{a_2=1}^{m/d}\dots\sum_{a_n=1}^{m/d}[gcd(a_1,a_2,\dots,a_n)=1](a_1*a_2*\dots*a_n)^k \\ =d^{kn}\sum_{a_1=1}^{m/d}\sum_{a_2=1}^{m/d}\dots\sum_{a_n=1}^{m/d}\sum_{j|gcd(a_1,a_2,\dots,a_n)}\mu(j)(a_1*a_2*\dots*a_n)^k \\ =d^{kn}\sum_{j=1}^{m/d}\mu(j)\ j^k\sum_{a_1=1}^{m/d/j}\sum_{a_2=1}^{m/d/j}\dots\sum_{a_n=1}^{m/d/j}(a_1*a_2*\dots*a_n)^k \\ =d^{kn}\sum_{j=1}^{m/d}\mu(j)\ j^k(\sum_{i=1}^{m/d/j}i^k)^n \\ \]這樣我們就可以預處理出\(1^k,2^k,3^k,\dots,m^k\),然後列舉\(\mathit j\)進行快速求解,時間複雜度:\(O(T*m*logk)\)。
觀察答案的計算式可以發現,\(\mathit n\)只存在於指數部分,於是我們可以利用尤拉降冪來將其降數量級。
尤拉降冪的公式為:
程式碼:
#include <iostream> #include <cstdio> #include <algorithm> #include <bits/stdc++.h> #define ALL(x) (x).begin(), (x).end() #define sz(a) int(a.size()) #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), '\0', sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl #define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c)) #define du2(a,b) scanf("%d %d",&(a),&(b)) #define du1(a) scanf("%d",&(a)); using namespace std; typedef long long ll; ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;} ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;} ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;} void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}} void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}} inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') { fh = -1; } c = getchar();} while (c >= '0' && c <= '9') { tmp = tmp * 10 + c - 48, c = getchar(); } return tmp * fh;} inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') { fh = -1; } c = getchar();} while (c >= '0' && c <= '9') { tmp = tmp * 10 + c - 48, c = getchar(); } return tmp * fh;} void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '\n' : ' ');}} void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '\n' : ' ');}} const int maxn = 100010; const int inf = 0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ #define DEBUG_Switch 0 bool vis[maxn]; long long prim[maxn], mu[maxn], sum[maxn], cnt; void get_mu(long long n) { mu[1] = 1; for (long long i = 2; i <= n; i++) { if (!vis[i]) {mu[i] = -1; prim[++cnt] = i;} for (long long j = 1; j <= cnt && i * prim[j] <= n; j++) { vis[i * prim[j]] = 1; if (i % prim[j] == 0) { break; } else { mu[i * prim[j]] = -mu[i]; } } } // for (long long i = 1; i <= n; i++) { sum[i] = sum[i - 1] + mu[i]; } } ll euler(ll n) //log(n)時間內求一個數的尤拉值 { ll ans = n; for (ll i = 2; i * i <= n; i++) { if (n % i == 0) { ans -= ans / i; while (n % i == 0) { n /= i; } } } if (n > 1) { ans -= ans / n; } return ans; } char s[maxn]; ll m, d, k; ll a[maxn]; int main() { #if DEBUG_Switch freopen("D:\\code\\input.txt", "r", stdin); #endif //freopen("D:\\code\\output.txt","w",stdout); get_mu(maxn - 1); ll mod = euler(59964251); ll base = 59964251ll; int t; t = readint(); while (t--) { scanf("%s %lld %lld %lld", s + 1, &m, &d, &k); int len = strlen(s + 1); ll n = 0; repd(i, 1, len) { n = n * 10 + (s[i] - '0'); if (n > mod) { n = n % mod + mod; } } repd(i, 1, m / d) { a[i] = powmod(i, k, base); sum[i] = (a[i] + sum[i - 1]) % base; } repd(i, 1, m / d) { sum[i] = powmod(sum[i], n, base); } ll ans = 0ll; repd(j, 1, m / d) { ans += mu[j] * sum[m / d / j] % base * powmod(a[j], n, base) % base; ans = (ans + base) % base; } ans = ans * powmod(d, k * n , base) % base; printf("%lld\n", ans ); } return 0; }