题面

T1

思路

考虑一下每个数会与其他位置的哪些数字遇到。显然每隔gcd(n,m,k)个数都会遇到一次。所以只要看一下将给出的所有数字全部对gcd(n,m,k)取模是否能包含从0到gcd(n,m,k) - 1的所有数就行了。

代码

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 100000 + 100;
ll read() {
    ll x = 0, f = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
int n,m,K,a[N],b[N],k[N];
int gcd(int x,int y) {
    return !y ? x : gcd(y,x % y);
}
namespace BF1 {
    
    void Main() {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        int aa = read();
        for(int i = 1;i <= aa;++i) a[read()] = 1;
        int bb = read();
        for(int i = 1;i <= bb;++i) b[read()] = 1;
        read();
        for(int i = 0;i <= 100000;++i) {
            int x1 = i % n, x2 = i % m;
            a[x1] = b[x2] = a[x1] | b[x2];
        }
        for(int i = 0;i < n;++i) {
            if(a[i] != 1) {
                puts("No");
                return;
            }
        }
        for(int i = 0;i < m;++i) {
            if(b[i] != 1) {
                puts("No");
                return;
            }
        }
        puts("Yes");
    }
}
namespace BF2 {
    int tmp[N * 5],js = 0;
    void Main() {
        js = 0;
        int mod = gcd(gcd(n,m),K);
        int aa = read();
        for(int i = 1;i <= aa;++i) tmp[++js] = read() % mod;
        int bb = read();
        for(int i = 1;i <= bb;++i) tmp[++js] = read() % mod;
        int kk = read();
        for(int i = 1;i <= kk;++i) tmp[++js] = read() % mod;
        int now = 0;
        sort(tmp + 1,tmp + js + 1); 
        tmp[0] = -1;
        int ans = 0;
        for(int i = 1;i <= js;++i) {
            if(tmp[i] == tmp[i-1]) continue;
            if(tmp[i] == tmp[i-1]+1) now++;
            else {
                ans = max(ans,now);
                now = 1;
            }
        }
        ans = max(ans,now);
        if(ans >= mod) puts("Yes");
        else puts("No");
        return;
    }
}
int main() {
    freopen("happy2.in","r",stdin);
    freopen("happy2.out","w",stdout);
    int T = read();
    while(T--) {
        n = read(),m = read(),K = read();
        if(n <= 100 && m <= 100 && K == 0) {
            BF1::Main();
            continue;
        }
        BF2::Main();
    }
    return 0;
}

T2

想了一会感觉不可做，直接55分暴力。

55分代码

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<cstring>
#include<algorithm>
#include<bitset>
using namespace std;
typedef long long ll;
const int N = 300000 + 100;
ll read() {
    ll x = 0, f = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
int n,p;
namespace BF1 {
    int du[N];
    void Main() {
        for(int i = 1;i <= n;++i) {
            int x = read(), y = read();
            du[x]++;
            du[y]++;
        }
        ll ans1 = 0;
        for(int i = 1;i <= n;++i)
            if(du[i] == 0) ans1++;
        cout<<(ll)n * (ll)(n - 1)/2 - (ans1 * (ans1 - 1) / 2);
        return;
    }
}
namespace BF2 {
    const int NN = 110;
    bitset <NN> tmp[NN];
    int ans = 0;
    inline int check(int x,int y) {
        bitset <NN> ls;
        ls = tmp[x] | tmp[y];
        return ls.count() >= p;
    }
    void Main() {
        for(int i = 1;i <= n;++i) {
            int x = read(),y = read();
            tmp[x][i] = 1;
            tmp[y][i] = 1;
        }
        for(int i = 1;i <= n;++i) {
            for(int j = i + 1;j <= n;++j) {
                ans += check(i,j);
            }
        }
        cout<<ans<<endl;
    }
}
int main() {
    freopen("suspect.in","r",stdin);
    freopen("suspect.out","w",stdout);
    n =  read(),p = read();
    if(p == 0) {
        cout<<((ll)n*(n-