Codeforces Edu Round 68 (Rated for Div. 2)

比较水的一场。

题目链接：https://codeforces.com/contest/1194

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A：

秒懂跟x没关系，答案就是2n。

B：

因为n*m<=4e5，统计每行每列点的数量后直接O(nm)暴力枚举维护最小值即可。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson curpos<<1
#define rson curpos<<1|1
/* namespace */
using namespace std;
/* header end */

int main() {
    int t; cin >> t;
    while (t--) {
        int n, m, ans = int_inf; cin >> n >> m;
        char s[n + 1][m + 1];
        int row[n + 1] = {0}, col[m + 1] = {0};
        rep1(i, 1, n) {
            scanf("%s", s[i] + 1);
            rep1(j, 1, m) {
                if (s[i][j] == '.') row[i]++, col[j]++;
            }
        }
        rep1(i, 1, n) {
            rep1(j, 1, m) {
                int k = row[i] + col[j];
                if (s[i][j] == '.') k--;
                ans = min(ans, k);
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}

C：

O(n)扫一遍就行了。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson curpos<<1
#define rson curpos<<1|1
/* namespace */
using namespace std;
/* header end */

const int maxn = 110;
int q;

int main() {
    scanf("%d", &q);
    while (q--) {
        string s, t, p; cin >> s >> t >> p;
        int len1 = s.size(), len2 = t.size(), len3 = p.size(), ans = 1;
        int a[26] = {0};
        for (auto c : p) a[c - 'a']++;
        int i = 0, j = 0;
        for (; i < len2; i++) {
            if (j < len1 && t[i] == s[j]) j++;
            else if (a[t[i] - 'a']) a[t[i] - 'a']--;
            else {
                ans = 0;
                break;
            }
        }
        if (j != len1) ans = 0;
        if (ans) puts("YES"); else puts("NO");
    }
    return 0;
}

D：

一看就是SG函数相关的题，打个表找规律发现循环节长度为3。简单推一下必胜必败点也可以猜出来。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson curpos<<1
#define rson curpos<<1|1
/* namespace */
using namespace std;
/* header end */

int t;

int main() {
    scanf("%d", &t);
    while (t--) {
        int n, k; scanf("%d%d", &n, &k);
        if (k % 3 == 0) {
            int p = n % (k + 1) + 1;
            if (p == k + 1) {
                puts("Alice");
                continue;
            } else {
                if (p % 3 == 1) puts("Bob");
                else puts("Alice");
            }
        } else {
            if (n % 3 == 0) puts("Bob");
            else puts("Alice");
        }
    }
    return 0;
}

E：

树状数组+扫描线。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson curpos<<1
#define rson curpos<<1|1
/* namespace */
using namespace std;
/* header end */

const int maxn = 1e4 + 10;
int n, m, tot1 = 0, tot2 = 0, c[maxn], vis[maxn];
ll ans = 0;

struct Vert {
    int x, y, xx;
    bool operator<(const Vert &rhs)const {
        return y < rhs.y;
    }
} v[maxn];

struct Hori {
    int x, y, yy;
    bool operator<(const Hori &rhs)const {
        return yy < rhs.yy;
    }
} h[maxn];

int lowbit(int x) {
    return x & -x;
}

int find(int x) {
    int ret = 0;
    while (x > 0) {
        ret += c[x];
        x -= lowbit(x);
    }
    return ret;
}

void add(int x, int val) {
    while (x <= 10001) {
        c[x] += val;
        x += lowbit(x);
    }
}

int main() {
    scanf("%d", &n);
    rep1(i, 1, n) {
        int x, y, xx, yy; scanf("%d%d%d%d", &x, &y, &xx, &yy);
        x += 5001, y += 5001, xx += 5001, yy += 5001;
        if (x == xx) h[++tot2] = (Hori) {
            x, min(y, yy), max(y, yy)
        };
        else v[++tot1] = (Vert) {
            min(x, xx), y, max(x, xx)
        };
    }
    sot(v, tot1); sot(h, tot2);
    sort(v + 1, v + 1 + tot1);
    sort(h + 1, h + 1 + tot2);
    rep1(i, 1, tot1) {
        rep0(j, 0, maxn) c[j] = 0;
        rep1(j, 1, tot2) {
            if (h[j].y <= v[i].y && h[j].yy >= v[i].y) {
                add(h[j].x, 1);
                vis[j] = 1;
            } else vis[j] = 0;
        }
        int k = 1;
        rep1(j, i + 1, tot1) {
            while (k <= tot2 && h[k].yy < v[j].y) {
                if (vis[k]) add(h[k].x, -1);
                k++;
            }
            int l = max(v[i].x, v[j].x) - 1, r = min(v[i].xx, v[j].xx);
            if (l < r) {
                int x = find(r) - find(l);
                ans += 1LL * x * (x - 1) / 2;
            }
        }
    }
    printf("%lld\n", ans);
    return 0;
}

F && G：

不会，溜了(