[20181107][模拟赛]

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-1) / 2);
        return 0;
    }
    if(p == 1) {
        BF1::Main();
        return 0;
    }
    if(n <= 100) {
        BF2::Main();
        return 0;
    }
    return 0;
}

std

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<cstdlib>
#define INF 100000000
using namespace std;
typedef long long LL;
struct Edge
{
    int from,to,pre;
}e[1000000];
int h[300005]={0},cou=0;
int c[300005],ed[300005];
void Addedge(int from,int to)
{
    cou++;
    e[cou]=((Edge){from,to,h[from]});
    h[from]=cou;
}

inline void update(int x)
{
    if(x==0)
    {
        c[0]++;
        return;
    }
    for(;x<=300000;x+=x&-x)
        c[x]++;
}
inline int get(int x)
{
    if(x==-1) return 0;
    int sum=0;
    for(;x;x-=x&-x)
        sum+=c[x];
    return sum+c[0];
}
int main()
{
    freopen("suspect.in","r",stdin);
    freopen("suspect.out","w",stdout);
    LL ans=0;
    int sum,i,n,p,a,b,v,j;
    cin>>n>>p;
    for(i=1;i<=n;i++)
    {
        scanf("%d%d",&a,&b);
        Addedge(a,b); Addedge(b,a);
        ed[a]++; ed[b]++;
    }
    for(i=1;i<=n;i++)
        update(ed[i]);
    for(i=1;i<=n;i++)
    {
        if(ed[i]>=p)
            ans+=n-1;
        else
        {
            sum=n-get(p-ed[i]-1);
            if(ed[i]>=p-ed[i]) sum--;
            for(j=h[i];j;j=e[j].pre)
            {
                v=e[j].to;
                if(ed[v]==p-ed[i]) sum--;
                ed[v]--;
            }
            for(j=h[i];j;j=e[j].pre)
            {
                v=e[j].to;
                ed[v]++;
            }
            ans+=sum;
        }
    }
    cout<<ans/2<<endl;
    return 0;
}

T3

80分思路

暴力分感觉都可做，然后就写了80分暴力。用莫队卡一下100分？？？会TLE吧(真麻烦，不想写)。

100分思路

如果这个题让着求区间出现奇数次的数的异或和就很简单了。所以我们可以对于询问先离线一下。按照右端点拍个序。然后用树状数组维护一下区间异或和。就是从1扫到n，同时将当前的数上一次出现的位置(在树状数组里)异或上这个数。然后查询就好了。

100分代码

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<cstdlib>
#include<string>
#include<bitset>
#include<iomanip>
#include<deque>
#include<utility>
#define INF 1000000000
#define fi first
#define se second
#define N 1000005
#define P 1000000007
#define debug(x) cerr<<#x<<"="<<x<<endl
#define MP(x,y) make_pair(x,y)
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
int c[N],now,sum,a[N],b[N],ans[N],nxt[N],n;
map<int,int> vis,pre;
vector<pii> Q[N];

void Add(int x,int w)
{
    for(;x<=n;x+=x&-x)
        c[x]^=w;
}

int Get(int x)
{
    int s=0;
    for(;x;x-=x&-x)
        s^=c[x];
    return s;
}

int main()
{
    int i,m,ql,qr,j;
    freopen("xor.in","r",stdin);
    freopen("xor.out","w",stdout);
    cin>>n;
    now=0;
    for(i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        vis[a[i]]++;
        nxt[pre[a[i]]]=i;
        pre[a[i]]=i;
        if(vis[a[i]]>1)
            now^=a[i];
        b[i]=now;
        //debug(b[i]);
    }
    cin>>m;
    for(i=1;i<=m;i++)
    {
        scanf("%d%d",&ql,&qr);
        Q[ql].push_back(MP(qr,i));
    }
    for(i=1;i<=n;i++)
    {
        //debug(sum);
        for(j=0;j<Q[i].size();j++)
            ans[Q[i][j].se]=Get(Q[i][j].fi)^b[Q[i][j].fi];
        ql=nxt[i];
        if(ql)
        {
            sum^=a[i];
            Add(ql,a[i]);
        }
    }
    for(i=1;i<=m;i++)
        printf("%d\n",ans[i]);
    return 0;
}