【LOJ】#3096. 「SNOI2019」數論

LOJ#3096. 「SNOI2019」數論

若是\(P > Q\)咱們把\(P\)和\(Q\)換一下，如今默認\(P < Q\)

這個時候每一個合法的\(a_i\)均可以直接落到\(Q\)中，由於\(a_{i} \equiv a_{i} \pmod Q\)這樣避免了麻煩

而後呢咱們發現每次把\((a_{i} + P) \% Q\)會走成一個圈，咱們就要求從\(a_{i}\)開始數\(\lfloor \frac{T - 1- a_{i}}{P} \rfloor + 1\)個圈裏\(b_{i}\)的總和

這樣的話能夠斷圈爲鏈，複製兩份就能夠快速查某一段開始走小於圈長步的總和了

這樣的圈一共有\(gcd(P,Q)\)個

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define ba 47
#define MAXN 1000005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int P,Q,n,m;
int A[MAXN],B[MAXN];
int sum[MAXN * 2],pos[MAXN * 2],tot;
bool vis[MAXN];
int64 ans,T;
int gcd(int a,int b) {
    return b == 0 ? a : gcd(b,a % b);
}
void Solve() {
    read(P);read(Q);read(n);read(m);read(T);
    int x;
    for(int i = 1 ; i <= n ; ++i) {
    read(x);A[x] = 1;
    }
    for(int i = 1 ; i <= m ; ++i) {
    read(x);B[x] = 1;
    }
    if(P > Q) {
    swap(P,Q);
    for(int i = 0 ; i <= 1000000 ; ++i) swap(A[i],B[i]);
    }
    int g = gcd(P,Q);
    for(int i = 0 ; i < g ; ++i) {
    for(int j = 1 ; j <= tot ; ++j) sum[j] = 0;
    tot = 0;
    int x = i;
    while(!vis[x]) {
        sum[tot + 1] = B[x] + sum[tot];
        pos[tot + 1] = x;
        ++tot;
        vis[x] = 1;
        x = (x + P) % Q;
    }
    int t = tot;
    for(int j = 1 ; j <= t ; ++j) {
        if(pos[j] < P && A[pos[j]]) {
        if(T - 1 - pos[j] >= 0) {
            int64 all = (T - 1 - pos[j]) / P + 1;
            int64 cnt = all / t;
            ans += cnt * sum[t];
            int rem = all % t;
            if(rem) {
            ans += sum[j + rem - 1] - sum[j - 1];
            }
        }
        }
        sum[tot + 1] = sum[tot] + B[x];++tot;
        x = (x + P) % Q;
    }
    }
    out(ans);enter;
}
int main(){
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}