联赛前第九阶段总结

联赛模拟测试31

A 异或

• 求出来每一位l到r每个数的0和1的个数，乘起来再乘这个位的值乘2加起来就是答案

Show Code

#include <cstdio>
#include <algorithm>

const int M = 1e9 + 7;
int T;

int main() {
    freopen("xor.in", "r", stdin);
    freopen("xor.out", "w", stdout);
    scanf("%d", &T);
    while (T--) {
        int l, r, ans = 0;
        scanf("%d%d", &l, &r);
        l, r++;
        for (int i = 1; i < r; i <<= 1) {
            int s = r / (i << 1) * i + std::max(r % (i << 1) - i, 0);
            s -= l / (i << 1) * i + std::max(l % (i << 1) - i, 0);
            if ((ans += 1LL * s * (r - l - s) % M * i % M) >= M) ans -= M;
        }
        printf("%d
", ans * 2 % M);
    }
    return 0;
}

B 取石子 (Unaccepted)

• 大力puts("Yes"),10分到手

C 优化 (Unaccepted)

• 瞎写了个DP，样例都过不去，0分

D 计数 (Unaccepted)

• 拿了M=0的20分，暴搜好像打挂了

联赛模拟测试30

阶段排名 10

A maze

• 想出来是s关于k是单增的了，但当时只想这个函数怎么求，没想到二分，就写了10分

Show Code

#include <queue>
#include <cstdio>

const int N = 105;

int read(int x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar())
        if (c == '-') f = -1;
    for (; c >='0' && c <='9'; c = getchar())
        x = x * 10 + c - '0';
    return x * f;
}

struct Edge {
    int next, t; bool d;
}e[N*N<<2];
int head[N*N], edc;

void Add(int x, int y, bool d) {
    e[++edc] = (Edge) {head[x], y, d};
    head[x] = edc;
}

int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};
int n, m, h[N][N], hc, s, t;
bool a[N][N], v[N*N];
double sd, d[N*N];

double Dij(double mid) {
    std::priority_queue< std::pair<double, int> > q;
    for (int i = 1; i <= n * m; ++i)
        d[i] = 1e9, v[i] = 0;
    q.push(std::make_pair(d[s] = 0, s));
    while (!q.empty()) {
        int x = q.top().second; q.pop();
        if (x == t) return d[x];
        if (v[x]) continue;
        v[x] = 1;
        for (int i = head[x]; i; i = e[i].next) {
            int y = e[i].t;
            double dis = e[i].d ? 1 : mid;
            if (d[y] > d[x] + dis) {
                d[y] = d[x] + dis;
                q.push(std::make_pair(-d[y], y));
            }
        }
    }
}

int main() {
    freopen("maze.in", "r", stdin);
    freopen("maze.out", "w", stdout);
    n = read(); m = read();
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            h[i][j] = ++hc;
    s = read(); s = h[s][read()];
    t = read(); t = h[t][read()];
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            a[i][j] = read();
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            if (a[i][j]) continue;
            for (int k = 0; k < 4; ++k) {
                int x = i + dx[k], y = j + dy[k];
                if (x < 1 || x > n || y < 1 || y > m || a[i][j]) continue;
                Add(h[i][j], h[x][y], k < 2);
            }
        }
    }
    scanf("%lf", &sd);
    double l = 0, r = sd;
    while (r - l > 1e-5) {
        double mid = (l + r) / 2;
        if (Dij(mid) >= sd) r = mid;
        else l = mid;
    }
    printf("%.3lf
", l);
    return 0;
}

B bird

• 写了个n^3暴力拿了50分

• 题目中说鸟每秒向负方向移动一个单位，但是鸟太多了，不好处理，所以就让人每秒向正方向移动一个单位

• f[i]表示最后一枪打到i位置最多能打到多少只鸟，显然是要从i-k之前的找最大值转移，但是这道题一只鸟可能会很长，但一只鸟又不能算两次，我考场就写了个bitset判断

• 考虑将鸟都算在最后一枪上，前面的都不算当前可以打的到的鸟，当一只鸟在i位置打不到就可以对后面的做贡献，就加进去

• 所以维护一颗支持区间加，区间查询最大值的线段树即可，每个叶子节点是f[i]-i位置可以打的到的鸟

Show Code

#include <cstdio>
#include <vector>
#define ls (rt << 1)
#define rs (rt << 1 | 1)

const int N = 5e5 + 5;

int read(int x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar())
        if (c == '-') f = -1;
    for (; c >='0' && c <='9'; c = getchar())
        x = x * 10 + c - '0';
    return x * f;
}

std::vector<int> a[N];
int n, k, m, s[N], ans, t[N<<2], tag[N<<2];

void Pushdown(int rt) {
    t[ls] += tag[rt]; tag[ls] += tag[rt];
    t[rs] += tag[rt]; tag[rs] += tag[rt];
    tag[rt] = 0;
}

void Change(int rt, int l, int r, int x, int y, int w) {
    if (x <= l && r <= y) return t[rt] += w, tag[rt] += w, void();
    if (tag[rt]) Pushdown(rt);
    int mid = l + r >> 1;
    if (x <= mid) Change(ls, l, mid, x, y, w);
    if (y > mid) Change(rs, mid + 1, r, x, y, w);
    t[rt] = std::max(t[ls], t[rs]);
}

int Ask(int rt, int l, int r, int x, int y) {
    if (x <= l && r <= y) return t[rt];
    if (tag[rt]) Pushdown(rt);
    int mid = l + r >> 1, ans = 0;
    if (x <= mid) ans = Ask(ls, l, mid, x, y);
    if (y > mid) ans = std::max(ans, Ask(rs, mid + 1, r, x, y));
    return ans;
}

int main() {
    freopen("bird.in", "r", stdin);
    freopen("bird.out", "w", stdout);
    m = read(); k = read();
    while (m--) {
        int l = std::max(0, read()) + 1, r = read() + 1;
        if (r < 1) continue;
        if (r > n) n = r;
        s[l]++, s[r+1]--;
        a[r].push_back(l);
    }
    for (int i = 1; i <= n; ++i) {
        s[i] += s[i-1];
        int f = i > k ? Ask(1, 1, n, 1, i - k) : 0;
        Change(1, 1, n, i, i, f);
        if ((f += s[i]) > ans) ans = f;
        for (int j = 0; j < a[i].size(); ++j)
            Change(1, 1, n, a[i][j], i, 1);
    }
    printf("%d
", ans);
    return 0;
}

C stone (Unaccepted)

• 30分暴力数组开小拿了20

D 椎 (Unaccepted)

• 写了个Treap，过了样例，但前一半WA，后一半RE

上午小测6

阶段排名 7

A 大新闻

• 打出nm暴力之后就考虑怎么优化

• 求第k大肯定是套什么数据结构了，我知道的就权值线段树可以，好像还有什么平衡树，可是我太菜了

• 插入删除得把整个数组左右移动，但是一个很好的性质就是都是在开头进行处理，那把序列反过来，每次在结尾处理不就好了吗

• 问题就转换成在序列最后加入或删除一个数，然后就区间第k小，

• 动态区间第k小不久是主席树板子么，就多了一个删除，删除就直接n--就没事了

• 需要注意区间被倒过来了，查询k小的区间就不是[l,r]，而是[n-r+1, n-l+1]

Show Code

#include <cstdio>
#include <cstring>
#include <algorithm>
#define ls(x) t[x].l
#define rs(x) t[x].r

const int N = 4e5 + 5;

int read(int x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar())
        if (c == '-') f = -1;
    for (; c >= '0' && c <= '9'; c = getchar())
        x = x * 10 + c - '0';
    return x * f;
}

struct tree {
    int s, l, r;
}t[N*31];
int n, m, a[N], root[N], trc;

void Change(int &rt, int l, int r, int x) {
    t[++trc] = t[rt]; rt = trc;
    t[rt].s++;
    if (l == r) return;
    int mid = l + r >> 1;
    if (x <= mid) Change(ls(rt), l, mid, x);
    else Change(rs(rt), mid + 1, r, x);
}

int Ask(int lrt, int rrt, int l, int r, int k) {
    if (l == r) return l;
    int mid = l + r >> 1, s = t[ls(rrt)].s - t[ls(lrt)].s;
    if (k <= s) return Ask(ls(lrt), ls(rrt), l, mid, k);
    else return Ask(rs(lrt), rs(rrt), mid + 1, r, k - s);
}

int main() {
    freopen("news.in", "r", stdin);
    freopen("news.out", "w", stdout);
    n = read(), m = read();
    for (int i = n; i >= 1; --i)
        a[i] = read();
    for (int i = 1; i <= n; ++i)
        Change(root[i] = root[i-1], 1, 1e9, a[i]);
    while (m--) {
        int od = read();
        if (od == 1) n--;
        else if (od == 2) n++, Change(root[n] = root[n-1], 1, 1e9, read());
        else {
            int r = n - read() + 1, l = n - read() + 1, k = read();
            printf("%d
", Ask(root[l-1], root[r], 1, 1e9, k));
        }
    }
    return 0;
}

B King's Inspection (Unaccepted)

• 暴力直接94

联赛模拟测试29(lrd day2)

阶段排名 10

A 串串香

• 最后检查的时候还在纠结要不要define int long long，后来没改，结果28行n乘的时候炸成90分了

• 前缀和后缀相等就转换成循环节问题，本来就有长度为m的循环节，然后让循环节尽可能小，就在m里再找循环节

Show Code

#include <cstdio>

typedef unsigned long long ull;
const int N = 1e6 + 5;
ull h[N], p[N];
long long ans, n, m;
int T;
char c[N];

ull Ask(int l, int r) {
    return h[r] - h[l-1] * p[r-l+1];
}

int main() {
    freopen("ccx.in", "r", stdin);
    freopen("ccx.out", "w", stdout);
    scanf("%d", &T);
    p[0] = 1;
    for (int i = 1; i < N; ++i)
        p[i] = p[i-1] * 233;
    while (T--) {
        scanf("%lld%lld%s", &n, &m, c + 1);
        for (int i = 1; i <= m; ++i)
            h[i] = h[i-1] * 233 + c[i];
        for (int i = 1; i * 2 <= m; ++i) {
            if (m % i) continue;
            if (Ask(1, m - i) == Ask(1 + i, m)) {
                n *= m / i; m = i;
                break;
            }
        }
        ans = (n - 1) * m;
        if (!ans) {
            ans = -1;
            for (int i = 1; i < m; ++i)
                if (Ask(1, i) == Ask(m - i + 1, m))
                    ans = i;
        }
        printf("%lld
", ans);
    }
    return 0;
}

B 糊涂图 (Unaccepted)

• 概率跳过

C 木叶下 (Unaccepted)

• 瞎打部分分拿35

D Tiny Counting

• 离散化后二分查找a[i]的时候把范围1～m写成1～n了，就暴成70分了

• 树状数组求出每个点前面大的，小的，后面大的，小的

• 先不考虑ab和cd有重复，那就是正序对数乘逆序对数

• 当a=c,a=d,b=c,b=d的时候这样会有重复，就把这几种情况减去就可以了

Show Code

#include <cstdio>
#include <cstring>
#include <algorithm>

const int N = 1e5 + 5;

int read(int x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar())
        if (c == '-') f = -1;
    for (; c >= '0' && c <= '9';c = getchar())
        x = x * 10 + c - '0';
    return x * f;
}

int n, m, a[N], b[N], b1[N], b2, s1[N], s2, tb[N], ts[N];
long long sb, ss, ans;

void Addb(int x) {
    for (; x; x -= x & -x)
        tb[x]++;
}

int Askb(int x, int ans = 0) {
    for (; x <= m; x += x & -x)
        ans += tb[x];
    return ans;
}

void Adds(int x) {
    for (; x <= m; x += x & -x)
        ts[x]++;
}

int Asks(int x, int ans = 0) {
    for (; x; x -= x & -x)
        ans += ts[x];
    return ans;
}

signed main() {
    freopen("a.in", "r", stdin);
    freopen("a.out", "w", stdout);
    n = read();
    for (int i = 1; i <= n; ++i)
        a[i] = b[i] = read();
    std::sort(b + 1, b + n + 1);
    m = std::unique(b + 1, b + n + 1) - b - 1;
    for (int i = 1; i <= n; ++i)
        a[i] = std::lower_bound(b + 1, b + m + 1, a[i]) - b;
    for (int i = 1; i <= n; ++i) {
        b1[i] = Askb(a[i] + 1); Addb(a[i]);
        s1[i] = Asks(a[i] - 1); Adds(a[i]);
        sb += b1[i];
        ss += s1[i];
        ans -= 1LL * b1[i] * s1[i];
    }
    ans += sb * ss;
    memset(tb, 0, sizeof(tb));
    memset(ts, 0, sizeof(ts));
    for (int i = n; i >= 1; --i) {
        b2 = Askb(a[i] + 1); Addb(a[i]);
        s2 = Asks(a[i] - 1); Adds(a[i]);
        ans -= 1LL * b2 * s2 + 1LL * b1[i] * b2 + 1LL * s1[i] * s2;
    }
    printf("%lld
", ans);
    return 0;
}

上午小测5

阶段排名 9

A 砍树

• 考场上写的二分，后来发现没有单调性，就写了个40分暴力，不过暴力和二分一个分

• 说实话并没有搞懂

• 问题等价于求一个最大的d,满足

[sum_{i=1}^{n}( left lceil frac{a_i}{d} ight ceil-a_i)leq k ]

• 移项整理，得

[dsum_{i=1}^{n}left lceil frac{a_i}{d} ight ceilleq k+sum_{i=1}^{n}a_i ]

• 注意到对于每个(a_i, left lceil a_i/d ight ceil),只有(sqrt{a_i})种不同的取值,因此 (sum_{i=1}^{n}left lceil a_i/d ight ceil) 只有(nsqrt{a_i})种不同的取值,在它的值确定之后,只需要简单的除法就可以求出d的最大值。因此把所有的不同的d的取值预处理出来排序,然后暴力计算,检验求出的d是否在这个取值所要求的d的范围内,并更新答案即可。

Show Code

#include <cstdio>
#include <algorithm>

typedef long long ll;

ll read(ll x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar())
        if (c == '-') f = -1;
    for (; c >='0' && c <='9'; c = getchar())
        x = x * 10 + c - '0';
    return x * f;
}

int n, m, a[105], b[32000000];
ll k, ans;

int main() {
    freopen("cut.in", "r", stdin);
    freopen("cut.out", "w", stdout);
    n = read(); k = read();
    for (int i = 1; i <= n; ++i) {
        a[i] = read();
        k += a[i];
        for (int j = 1; j * j <= a[i]; ++j)
            b[++m] = j, b[++m] = (a[i] - 1) / j + 1;
    }
    std::sort(b + 1, b + m + 1);
    m = std::unique(b + 1, b + m + 1) - b - 1;
    for (int i = 1; i <= m; ++i) {
        ll x = 0;
        for (int j = 1; j <= n; ++j)
            x += (a[j] - 1) / b[i] + 1;
        x = k / x;
        if (x >= b[i]) ans = x;
    }
    printf("%lld
", ans);
    return 0;
}

B 超级树 (Unaccepted)

• 暴搜拿了10分。

联赛模拟测试28(lrd day1)

阶段排名 9

A 位运算

• 考场上inf的判断挂掉了

&	|	^	组成情况
0	0	0	0 0
0	1	1	0 1 / 1 0
1	1	0	1 1

• 根据上表分类讨论一下就行了（下面是修改后的考场代码，懒得重写了）

Show Code

#include <cstdio>

typedef long long ll;

int read(int x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar())
        if (c == '-') f = -1;
    for (; c >= '0' && c <= '9'; c = getchar())
        x = x * 10 + c - '0';
    return x * f;
}

ll cnt = 0;

ll Pow(ll a, ll k, ll ans = 1) {
    for (; k; k >>= 1, a *= a) 
        if (k & 1) ans *= a;
    return ans;
}

int main() {
    freopen("bit.in", "r", stdin);
    freopen("bit.out", "w", stdout);
    int T = read();
    while (T--) {
        int a = read(), b = read(), c = read(), mi = a == -1 ? 0 : a; cnt = 0;
        if (a != -1 && c != -1 && (a & c)) puts("0");
        else if (b == -1 && (a == -1 || c == -1)) puts("inf");
        else if ((a != -1 && (a | b) != b) || (c != -1 && (b | c) != b))
                puts("0");
        else if (a != -1 && c != -1) {
            for (int i = 0; i < 30; ++i)
                if (!(a & (1 << i)) && (c & (1 << i))) cnt++;
            printf("%lld
", Pow(2, cnt));
        }
        else if (c != -1) {
            for (int i = 0; i < 30; ++i) {
                if (a != -1 && !(a & (1 << i)) && (b & (1 << i)) && !(c & (1 << i))) { 
                    cnt = -1; break;
                }
                if ((b & (1 << i)) && (c & (1 << i))) cnt++;
            }
            printf("%lld
", cnt == -1 ? 0 : Pow(2, cnt));
        }
        else if (a != -1) {
            for (int i = 0; i < 30; ++i)
                if (!(a & (1 << i)) && (b & (1 << i))) cnt++;
            printf("%lld
", Pow(2, cnt));
        }
        else if (a == -1 && c == -1) {
            for (int x = b; x; x -= x & -x) cnt++;
            printf("%lld
", Pow(3, cnt));
        }
    }
    return 0;
}

B 集合论

• 开个桶，加减就用tag标记，取交集就修改0点模拟清空，挺好写的

Show Code

#include <cstdio>

const int N = 1.1e6;

int read(int x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar())
        if (c == '-') f = -1;
    for (; c >= '0' && c <= '9'; c = getchar())
        x = x * 10 + c - '0';
    return x * f;
}

int b[N+N+5], d, z, siz;
long long s;

int main() {
    freopen("jihe.in", "r", stdin);
    freopen("jihe.out", "w", stdout);
    int n = read();
    for (int i = 1; i <= n; ++i) {
        int od = read();
        if (od == 1) {
            int m = read();
            while (m--) {
                int x = read();
                if (b[x-d+N] <= z) 
                    b[x-d+N] = i, s += x, siz++;
            }
        }
        else if (od == 2) {
            int m = read(); s = siz = 0;
            while (m--) {
                int x = read();
                if (b[x-d+N] > z)
                    b[x-d+N] = i, s += x, siz++;
            }
            z = i - 1;
        }
        else if (od == 3) d++, s += siz;
        else if (od == 4) d--, s -= siz; 
        printf("%lld
", s);
    }
    return 0;
}

C 连连看 (Unaccepted)

• 小数据暴力，k=0特判，拿了20分

D Huge Counting (Unaccepted)

• 一分暴力没取模，然后就爆0了

上午小测4

阶段排名 1

第二次AK！又是一道很难但数据极其水的题让我骗了100分

A Travel

• 考了一个半小时我才看出来，直接离线做，把边和询问按便权排序，双指针扫一遍，并查集维护加边

Show Code

#include <cstdio>
#include <algorithm>

const int N = 1e5 + 5;

int read(int x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar())
        if (c == '-') f = -1;
    for (; c >= '0' && c <= '9'; c = getchar())
        x = x * 10 + c - '0';
    return x * f;
}

struct Edge {
    int x, y, d;
    bool operator < (const Edge &b) const {
        return d < b.d;
    }
}e[N<<1];
struct Node {
    int x, w, id;
    bool operator < (const Node &b) const {
        return w < b.w;
    }
}a[N];
int n, m, q, f[N], siz[N], ans[N];

int Find(int x) {
    return x == f[x] ? x : (f[x] = Find(f[x]));
}

int main() {
    freopen("travel.in", "r", stdin);
    freopen("travel.out", "w", stdout);
    n = read(); m = read(); q = read();
    for (int i = 1; i <= n; ++i)
        f[i] = i, siz[i] = 1;
    for (int i = 1; i <= m; ++i)
        e[i] = (Edge) {read(), read(), read()};
    for (int i = 1; i <= q; ++i)
        a[i] = (Node) {read(), read(), i};
    std::sort(e + 1, e + m + 1);
    std::sort(a + 1, a + q + 1);
    for (int i = 1, j = 1; i <= q; ++i) {
        for (; j <= m && e[j].d <= a[i].w; ++j) {
            int x = Find(e[j].x), y = Find(e[j].y);
            if (x == y) continue;
            f[x] = y;
            siz[y] += siz[x];
        }
        ans[a[i].id] = siz[Find(a[i].x)];
    }
    for (int i = 1; i <= q; ++i)
        printf("%d
", ans[i]);
    return 0;
}

B Balanced Diet (Unaccepted)

• 考场上就写了一个O(ans*n)的写法，由于数据太水A掉了，在darkbzoj上搞到数据评测后是78分