杭电多校2023 第三场

1005

直接dp即可

#include <bits/stdc++.h>
using namespace std;
int dp[5005][5005];
int N;
int a[5005];
const int MOD = 1e9+7;
int main(){
    int T;
    cin >> T;
    while (T--){
        int N;
        memset(dp,0,sizeof(dp));
        dp[1][1] = 1;
        scanf("%d",&N);
        for (int i = 1 ; i <= N ; i ++){
            scanf("%d",&a[i]);
        }
        sort(a+1,a+N+1);
        for (int i = 2 ; i <= N ; i ++)
            if (a[i] == a[i-1]) dp[i][1] = dp[i-1][1];
            else dp[i][1] = dp[i-1][1]+1;
        for (int i = 2 ; i <= N ; i++)
            for (int j = 2 ; j <= i ; j ++){
                if (i == j) dp[i][j] = dp[i][j-1];
                else if (a[i] == a[i-1]) dp[i][j] = dp[i-1][j];
                else dp[i][j] = (dp[i][j-1]+dp[i-1][j])%MOD;
        }
        for (int i = 1 ; i <= N ; i ++)
            printf("%d\n",dp[N][i]);
    }
    return 0;
}

1011

队友写的，似乎是个大模拟？

#include <bits/stdc++.h>

#define x first
#define y second

using namespace std;

typedef long long LL;
typedef pair< int, int > PII;

const int N = 55;

int n, z;
char s[N][N];

int read()
{
    int res = 0, w = 1;
    char ch = getchar();
    while (ch != '-' && !isdigit(ch))
        ch = getchar();
    if (ch == '-')
        w = -1, ch = getchar();
    while (isdigit(ch))
        res = res * 10 + ch - '0', ch = getchar();
    return res * w;
}

void solve()
{
    scanf("%d%d", &n, &z);
    memset(s, 0, sizeof 0);
    for (int i = 1; i <= n; i++)
        scanf("%s", s[i] + 1);
        
    //    cout << "---" << endl;
    // for (int i = 1; i <= n; i++)
    // {
    //     for (int j = 1; j <= n; j++)
    //         cout << s[i][j];
    //     cout << endl;
    // }

    if ((n * z) % 100)
    {
        cout << "error" << endl;
        return;
    }

    if (z == 100)
    {
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
                cout << s[i][j];
            cout << endl;
        }
        return;
    }

    if (z == 125)
    {
        for (int i = 1; i <= n; i += 4)
            for (int j = 1; j <= n; j += 4)
            {
                char ch = s[i][j];
                for (int u = i; u < i + 4; u++)
                    for (int v = j; v < j + 4; v++)
                        if (s[u][v] != ch)
                        {
                            cout << "error" << endl;
                            return;
                        }
            }

        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j += 4)
            {
                for (int k = 1; k <= 5; k++)
                    cout << s[i][j];
            }
            cout << endl;

            if (i % 4 == 1)
            {
                for (int j = 1; j <= n; j += 4)
                {
                    for (int k = 1; k <= 5; k++)
                        cout << s[i][j];
                }
                cout << endl;
            }
        }
    }

    if (z == 150)
    {
        for (int i = 1; i <= n; i += 2)
            for (int j = 1; j <= n; j += 2)
            {
                char ch = s[i][j];

                for (int u = i; u < i + 2; u++)
                    for (int v = j; v < j + 2; v++)
                        if (s[u][v] != ch)
                        {
                            cout << "error" << endl;
                            return;
                        }
            }

        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j += 2)
            {
                for (int k = 1; k <= 3; k++)
                    cout << s[i][j];
            }
            cout << endl;

            if (i % 2 == 1)
            {
                for (int j = 1; j <= n; j += 2)
                {
                    for (int k = 1; k <= 3; k++)
                        cout << s[i][j];
                }
                cout << endl;
            }
        }
        return;
    }

    if (z == 175)
    {
        for (int i = 1; i <= n; i += 4)
            for (int j = 1; j <= n; j += 4)
            {
                char ch = s[i][j];

                for (int u = i; u < i + 4; u++)
                    for (int v = j; v < j + 4; v++)
                        if (s[u][v] != ch)
                        {
                            cout << "error" << endl;
                            return;
                        }
            }

        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j += 4)
            {
                for (int k = 1; k <= 7; k++)
                    cout << s[i][j];
            }
            cout << endl;

            if (i % 4 == 1)
            {
                for (int j = 1; j <= n; j += 4)
                {
                    for (int k = 1; k <= 7; k++)
                        cout << s[i][j];
                }
                cout << endl;
                for (int j = 1; j <= n; j += 4)
                {
                    for (int k = 1; k <= 7; k++)
                        cout << s[i][j];
                }
                cout << endl;
                for (int j = 1; j <= n; j += 4)
                {
                    for (int k = 1; k <= 7; k++)
                        cout << s[i][j];
                }
                cout << endl;
            }
        }
        return;
    }

    if (z == 200)
    {
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                for (int k = 1; k <= 2; k++)
                    cout << s[i][j];
            }
            cout << endl;

            for (int j = 1; j <= n; j++)
            {
                for (int k = 1; k <= 2; k++)
                    cout << s[i][j];
            }
            cout << endl;
        }
    }
}

int main()
{
    int T = 1;
    scanf("%d", &T);
    while (T--)
    {
        solve();
    }

    return 0;
}

1012

首先，可以把加的过程变成减的过程

然后就会发现这东西和gcd的那个更相减损术非常类似

于是就会考虑在辗转相除法的过程中，记录一下每个数对的情况所能达到的情况，并且排个序

每次查询的时候，就二分一下它在路径上的位置，算一个后缀个数即可。

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = 50010;

struct node {
    ll x, y;    
    bool operator < (const node &g) const {
        return x == g.x ? y < g.y : x < g.x;
    }
} ;

map<ll, int> t;

map<node, vector<ll> > pre;

int T, n, q;

ll a[N], b[N];

void gcd(ll x, ll y) {
    if (!x || !y) return ;
    if (x < y) {
        pre[(node){x, y % x}].push_back(y);
        gcd(x, y % x);
    }
    else {
        pre[(node){x % y, y}].push_back(x);
        gcd(x % y, y);
    }
}

int main() {
    scanf("%d", &T);
    for ( ; T; T--) {
        scanf("%d%d", &n, &q);
        t.clear(), pre.clear();
        for (int i = 1; i <= n; i++) {
            scanf("%lld%lld", &a[i], &b[i]);
            if (a[i] == b[i]) {
                t[a[i]]++;
                continue;    
            }
            gcd(a[i], b[i]);
            if (a[i] < b[i]) pre[(node){a[i], b[i]}].push_back(a[i]);    
            else pre[(node){a[i], b[i]}].push_back(b[i]);    
        }
        for (auto &[u,v]:pre){
            sort(v.begin(),v.end());
        }
        for (int i = 1; i <= q; i++) {
            ll c, d; scanf("%lld%lld", &c, &d);
            int cnt = 0;
            if (c == d) cnt = t[c];
            if (c <= d) {
                node g = (node){c, d % c};
                int siz = pre[g].size();
                int l = lower_bound(pre[g].begin(),pre[g].end(),d)-pre[g].begin();
                /*while (l < r) {
                    int mid = (l + r) >> 1;
                    if (now[mid] < d) l = mid + 1;
                    else r = mid;
                }*/
                cnt += siz - l;
            }
            if (c >= d) {
                node g = (node){c % d, d};
                int siz = pre[g].size();
                /*while (l < r) {
                    int mid = (l + r) >> 1;
                    if (now[mid] < c) l = mid + 1;
                    else r = mid;
                }*/
                int l = lower_bound(pre[g].begin(),pre[g].end(),c)-pre[g].begin();
                cnt += siz - l;
            }
            printf("%d\n", cnt);
        }
    }
    return 0;
}

1004

因为是全随机的数据

所以考虑每一个可能的$\delta X$和 $\delta Y$

随机抽取$100$个可能的值，然后考虑它是否合法

不合法就退出

因为数据全随机，所以其实不合法的概率非常高

#include <bits/stdc++.h>
#define pii pair<int,int>
using namespace std;
int T;
int N;
pii a[200005];
map<pii,int> cnt;
int Random(int MOD){
    return 1ll*rand() * 1ll*rand()%MOD*1ll*rand()%MOD;
}
int main(){
    srand(time(NULL));
    cin >> T;
    while (T--){
        cin >> N;
        cnt.clear(); 
        for (int i = 1 ; i <= 2 * N ; i ++){
            cin >> a[i].first >> a[i].second;
            cnt[a[i]] ++;
        }
        vector<pii> ans;
        for (int i = 2 ; i <= 2 * N ; i ++){
            cnt[a[1]] --;
            cnt[a[i]] --;
            int deltaX = a[i].first - a[1].first,deltaY = a[i].second - a[1].second;
            int TT = 1000;
            bool flag = false;
            while (TT --){
                int x = Random(2*N);
                x++;
                if ( x ==1 || x == i) continue;
                //cout << x << endl;
                pii New = {a[x].first + deltaX,a[x].second + deltaY};
                pii New1 ={a[x].first - deltaX,a[x].second - deltaY};
                if (cnt[New] + cnt[New1] < cnt[a[x]]){
                    flag = true;
                    break;
                }
            }
            if (!flag) {
                ans.push_back({deltaX,deltaY});
                ans.push_back({-deltaX,-deltaY});
            }
            cnt[a[1]] ++;
            cnt[a[i]] ++;
        }
        bool flag = false;
        for (int i = 1 ; i <= 2 * N ; i ++)
            if (cnt[a[i]] %2 != 0){
                flag = true;
                break;
            }
        if (flag == false) ans.push_back({0,0});
        sort(ans.begin(),ans.end());
        ans.erase(unique(ans.begin(),ans.end()),ans.end());
        int cntt = 0;
        for (auto v : ans){
            if (flag == true && v.first == 0 && v.second == 0) continue;
            cntt ++;
        }
        cout << cntt << '\n';
        for (auto v : ans){
            if (flag == true && v.first == 0 && v.second == 0) continue;
            cout << v.first << " " << v.second << '\n';
        }
    }
    return 0;
}