June Challenge 2018 Division 2

Naive Chef

暴力

#include <bits/stdc++.h>
using namespace std;
 
int main() {
    int T;
    scanf("%d", &T);
    while(T--){
        int N, A, B, x, ca = 0, cb = 0;
        scanf("%d %d %d", &N, &A, &B);
        for(int i = 1; i <= N; ++i) {
            scanf("%d", &x);
            if(x == A) ca++;
            if(x == B) cb++;
        }
        printf("%f
", 1.0 * ca * cb / N / N);
    }
    return 0;
} 

Binary Shuffle

每次可以加一个或者减到剩一个 特判几个点

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
 
int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        LL A, B, ca = 0, cb = 0;
        scanf("%lld %lld", &A, &B);
        if(A == B) {puts("0"); continue;}
        if(B == 0) {puts("-1"); continue;}
        B--;
        for (int i = 0; i <= 62; ++i) {
            if ((1LL << i) & A) ca++;
            if ((1LL << i) & B) cb++;
        }
        if(B == 0) {puts(A == 0 ? "1" : "-1"); continue;}
        if(ca == cb) puts("1");
        else if(ca > cb) puts("2");
        else printf("%d
", cb - ca + 1);
    }
    return 0;
} 

Vision

二分 我连三维叉积都不会了

using namespace std;
const double eps = 1e-9;
 
double sqr(double x) {return x * x;}
double dis(double x1, double y1, double z1, double x2, double y2, double z2) {
    return sqrt(sqr(x1 - x2) + sqr(y1 - y2) + sqr(z1 - z2));
}
double cross(double x1, double y1, double z1, double x2, double y2, double z2) {
    return sqrt(sqr(y1 * z2 - z1 * y2) + sqr(z1 * x2 - x1 * z2) + sqr(x1 * y2 - y1 * x2));
}
 
int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        double Px, Py, Pz;
        scanf("%lf %lf %lf", &Px, &Py, &Pz);
        double Qx, Qy, Qz;
        scanf("%lf %lf %lf", &Qx, &Qy, &Qz);
        double dx, dy, dz;
        scanf("%lf %lf %lf", &dx, &dy, &dz);
        double cx, cy, cz, r;
        scanf("%lf %lf %lf %lf", &cx, &cy, &cz, &r);
        double ans = 1e10, tmp = 1e10;
        while(tmp > eps) {
            double M = ans - tmp;
            double xx = Qx + dx * M;
            double yy = Qy + dy * M;
            double zz = Qz + dz * M;
            double PQ = dis(Px, Py, Pz, xx, yy, zz);
            double d = fabs(cross(Px - cx, Py - cy, Pz - cz, xx - cx, yy - cy, zz - cz)) / PQ;
            if(d >= r) ans = M;
            tmp /= 2;
        }
        printf("%.8f
", ans);
    }
    return 0;
} 

Sheokand and String

字典树

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
string S[maxn], P[maxn], ans[maxn];
int R[maxn], id[maxn];
char str[22];
 
bool cmp(int i, int j) {
    return R[i] < R[j];
}
 
int cnt, nxt[maxn][26], val[maxn];
void INS(int x) {
    int u = 0;
    for (int i = 0; i < S[x].length(); ++i) {
        if (nxt[u][S[x][i] - 'a']) u = nxt[u][S[x][i] - 'a'];
        else u = nxt[u][S[x][i] - 'a'] = ++cnt;
    }
    val[u] = 1;
}
string GET(int x) {
    string ret = "";
    int u = 0, flag = 0;
    for (int i = 0; i < P[x].length(); ++i) {
        if (!flag && nxt[u][P[x][i] - 'a']) ret += P[x][i], u = nxt[u][P[x][i] - 'a'];
        else {
            if (val[u]) return ret;
            flag = 1;
        }
        if (flag)
            for (int j = 0; j < 26; ++j) {
                if (nxt[u][j]) {
                    u = nxt[u][j];
                    ret += 'a' + j;
                    if (val[u]) return ret;
                    break;
                }
            }
    }
    if (val[u]) return ret;
    while (1) {
        for (int j = 0; j < 26; ++j) {
            if (nxt[u][j]) {
                u = nxt[u][j];
                ret += 'a' + j;
                if (val[u]) return ret;
                break;
            }
        }
    }
}
 
int main(){
    int N, Q;
    scanf("%d", &N);
    for(int i = 1; i <= N; ++i){
        scanf("%s", str);
        S[i] = string(str);
    }
    scanf("%d", &Q);
    for(int i = 1; i <= Q; ++i){
        scanf("%d %s", R + i, str);
        P[i] = string(str);
        id[i] = i;
    }
    sort(id + 1, id + 1 + Q, cmp);
    int p = 0;
    for(int i = 1; i <= Q; ++i) {
        int x = id[i];
        while(p + 1 <= R[x]) INS(++p);
        ans[x] = GET(x);
    }
    for(int i = 1; i <= Q; ++i) printf("%s
", ans[i].c_str());
    return 0;
} 

Two Flowers

枚举一个颜色的块 然后用并查集维护这个颜色以外的连通性 每条相邻边合并一次 所以是On的

#include <bits/stdc++.h>
using namespace std;
int ans, a[2222][2222], id[2222][2222];
 
typedef pair<int, int> pii;
map< int, vector<pii> > M;
map< int, vector<pii> > :: iterator it;
 
int r[4444444];
int fa[4444444];
stack<pii> FA, R;
int Find(int x) {
    return x == fa[x] ? x : Find(fa[x]);
}
void Union(int x, int y) {
    x = Find(x), y = Find(y);
    if(x == y) return;
    if(r[x] < r[y]) swap(x, y);
    FA.push(pii(y, fa[y]));
    R.push(pii(x, r[x]));
    fa[y] = x, r[x] += r[y];
}
 
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
void dfs(int i, int j, int x) {
    id[i][j] = x;
    r[x]++;
    ans = max(ans, r[x]);
    for (int o = 0; o < 4; ++o) {
        int nx = i + dx[o], ny = j + dy[o];
        if (a[nx][ny] == a[i][j] && !id[nx][ny]) dfs(nx, ny, x);
    }
}
 
int cnt, cnt2, vis[2222][2222];
map<int, int> mp;
void dfs1(int i, int j) {
    vis[i][j] = 1;
    for (int o = 0; o < 4; ++o) {
        int nx = i + dx[o], ny = j + dy[o];
        if (a[nx][ny] && a[nx][ny] != a[i][j]) {
            if (mp.find(a[nx][ny]) == mp.end())
                mp[a[nx][ny]] = ++cnt2 + cnt, fa[cnt + cnt2] = cnt + cnt2, r[cnt + cnt2] = r[id[i][j]];
            Union(id[nx][ny], mp[a[nx][ny]]);
            ans = max(ans, r[Find(id[nx][ny])]);
        }
        if (a[nx][ny] == a[i][j] && !vis[nx][ny]) dfs1(nx, ny);
    }
}
 
void cln(){
    while(!FA.empty()) {
        pii x = FA.top(); FA.pop();
        fa[x.first] = x.second;
    }
    while(!R.empty()) {
        pii x = R.top(); R.pop();
        r[x.first] = x.second;
    }
}
 
int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j)
            scanf("%d", &a[i][j]);
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= m; ++j) {
            if(id[i][j]) continue;
            dfs(i, j, ++cnt), fa[cnt] = cnt;
            M[a[i][j]].push_back(pii(i, j));
        }
    }
    for(it = M.begin(); it != M.end(); it++) {
        cnt2 = 0;
        vector<pii> & v = (*it).second;
        for(int i = 0; i < v.size(); ++i) {
            int x = v[i].first, y = v[i].second;
            mp.clear(), dfs1(x, y);
        }
        cln();
    }
    printf("%d
", ans);
    return 0;
} 

Ways to Work

考虑${d_i-i+1}$这个序列，每次可以加任意非负整数或者减一，枚举C的因子作为$d_{n}$然后倒着往前贪心，如果能+1就+1，否则减到最大的能整除的因子

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
vector<int> fac, ans;
 
int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        int N, C;
        scanf("%d %d", &N, &C);
        fac.clear();
        for(int i = 1; i <= C / i; ++i) {
            if(C % i == 0) {
                fac.push_back(i);
                if(i != C / i) fac.push_back(C / i);
            }
        }
        sort(fac.begin(), fac.end());
        for(int i = 0; i < fac.size(); ++i) {
            ans.clear();
            int cnt = 0, tmp = C;
            for(int j = i; ; ) {
                while (tmp % fac[j] != 0) j--;
                tmp /= fac[j], cnt++, ans.push_back(fac[j]);
                if(cnt == N || tmp == 1) break;
                if(tmp % (fac[j] + 1) == 0) j++;
            }
            if(tmp == 1) break;
        }
        for(int i = ans.size() + 1; i <= N; ++i) ans.push_back(1);
        for(int i = 0; i < N; ++i) printf("%d%c", ans[N-1-i] + i, i == N - 1 ? '
' : ' ');
    }
    return 0;
} 

Expected Buildings

只要求N段a的和，利用矩阵快速幂求前缀和减一下就可以了，又是向量乘……

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL mod = 163577857;
int p[50005], a[105], c[105];
int N, h, x, K;
 
struct Matrix {
    LL a[105][105];
    void init() {
        memset(a, 0, sizeof(a));
        for (int i = 0; i < 105; ++i) {
            a[i][i] = 1;
        }
    }
} P[30];
 
Matrix mul(Matrix a, Matrix b) {
    Matrix ans;
    memset(ans.a, 0, sizeof(ans.a));
    for (int i = 0; i < 105; ++i) {
        for (int k = 0; k < 105; ++k) {
            if (a.a[i][k] != 0)
                for (int j = 0; j < 105; ++j) {
                    ans.a[i][j] = (ans.a[i][j] + a.a[i][k] * b.a[k][j]) % mod;
                }
        }
    }
    return ans;
}
 
LL sum[105], b[105], cpy[105];
LL cal(int o) {
    if(o <= K) return sum[o];
    o -= K;
    for(int i = 1; i <= K; ++i) b[i] = a[1+K-i];
    b[K+1] = sum[K];
    for(int i = 0; i < 30; ++i) {
        if(o & (1 << i)) {
            memset(cpy, 0, sizeof(cpy));
            for(int j = 1; j <= K + 1; ++j)
                for(int k = 1; k <= K + 1; ++k)
                    cpy[j] = (cpy[j] + P[i].a[j][k] * b[k]) % mod;
            memcpy(b, cpy, sizeof(b));
        }
    }
    return b[K+1];
}
 
LL fp(LL a, int b) {
    LL ret = 1;
    while (b) {
        if (b & 1) ret = ret * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ret;
}
 
int main() {
    scanf("%d %d %d %d", &N, &h, &x, &K);
    for(int i = 1; i <= N; ++i) scanf("%d", p + i);
    for(int i = 1; i <= K; ++i) scanf("%d", a + i), sum[i] = (sum[i-1] + a[i]) % mod;
    for(int i = 1; i <= K; ++i) scanf("%d", c + i);
    for(int i = 1; i <= K; ++i) P[0].a[1][i] = c[i];
    for(int i = 2; i <= K; ++i) P[0].a[i][i-1] = 1;
    for(int i = 1; i <= K; ++i) P[0].a[K+1][i] = c[i];
    P[0].a[K+1][K+1] = 1;
    for(int i = 1; i < 30; ++i) P[i] = mul(P[i-1], P[i-1]);
    LL ans = 0;
    for(int i = 1; i <= N; ++i) {
        if(p[i] < x) ans = (ans + cal(p[i]) + cal(h) - cal(h - x + p[i]) + mod) % mod;
        else ans = (ans + cal(p[i]) - cal(p[i] - x) + mod) % mod;
    }
    printf("%lld
", ans * fp(cal(h), mod - 2) % mod);
    return 0;
} 

Warehouseman (Challenge)