Codeforces Round #445

ACM ICPC

每个队伍必须是3个人

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
using namespace std;
int cmp(const void * x, const void * y) {
    //x < y
    return (*((double *)(x))) > (*((double *)(y))) ? 1 : -1;
}
int a[6];
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    for (int i = 0; i < 6; i++) scanf("%d", &a[i]);
    int sum = 0;
    for (int i = 0; i < 6; i++) sum += a[i];
    bool f = false;
    for (int i = 0; i < 6; i++) {
        for (int j = 0; j < 6; j++) {
            for (int k = 0; k < 6; k++) {
                if (i != j && j != k && i != k && (a[i] + a[j] + a[k]) * 2 == sum)f = true;
            }
        }
    }
    if (f) printf("Yes
");
    else printf("No
");
    return 0;
}

Vlad and Cafes

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
using namespace std;
int cmp(const void * x, const void * y) {
    //x < y
    return (*((double *)(x))) > (*((double *)(y))) ? 1 : -1;
}
int a[200005];
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    int n, x, ans;
    scanf("%d", &n);
    memset(a, -1, sizeof(a));
    for (int i = 0; i < n; i++) {
        scanf("%d", &x);
        a[x] = i;
        ans = x;
    }
    for (int i = 0; i < 200005; i++) {
        if (a[i] != -1 && a[i] < a[ans]) ans = i;
    }
    printf("%d
", ans);
    return 0;
}

Petya and Catacombs

记录之前每个房间的最近访问时间，对每一个数字如果能用之前的房间来凑就先凑，凑不了就增加一个新房间。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
using namespace std;
int cmp(const void * x, const void * y) {
    //x < y
    return (*((double *)(x))) > (*((double *)(y))) ? 1 : -1;
}
int a[200005];
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    int ans = 1;
    a[0] = 1;
    int n, x;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &x);
        if (a[x] > 0) {
            a[x]--;
            a[i]++;
        } else {
            ans++;
            a[i]++;
        }
    }
    printf("%d
", ans);
    return 0;
}

Restoration of string

• If some string is the most frequent then all its substrings are the most frequent too.
• If string ab or similar is the most frequent then letter a is always followed by letter b and b always follow a.
• Let's consider directed graph on letters where edge a → b exists only if ab is the most frequent. If there is cycle in such graph then good string doesn't exist.
• So such graph can be represented as several non-intersecting paths. All strings which correspond to paths must occur in non-empty good string. So if we print them in lexicographical order then we will get the answer.

有3个判定条件：每个点入度不超过1；每个点出度不超过1；不能存在环。

按字典序输出所有路径

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
using std::vector;
using std::sort;
int cmp(const void * x, const void * y) {
    //x < y
    return (*((double *)(x))) > (*((double *)(y))) ? 1 : -1;
}
bool v[300], g[300];
int next[300], pre[300];
char str[100005];
bool dfs(int x) {
    if (g[x]) return false;
    g[x] = true;
    if (next[x] == 0) return true;
    return dfs(next[x]);
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    int n, len;
    while (scanf("%d", &n) != EOF) {
        memset(next, 0, sizeof(next));
        memset(pre, 0, sizeof(pre));
        memset(v, false, sizeof(v));
        bool flag = true;
        for (int i = 0; i < n; i++) {
            scanf("%s", str);
            len = strlen(str);
            for (int j = 0; j < len - 1; j++) {
                v[str[j]] = true;
                if (next[str[j]] == 0) next[str[j]] = str[j + 1];
                else if (next[str[j]] != str[j + 1]) flag = false;
                if (pre[str[j + 1]] == 0) pre[str[j + 1]] = str[j];
                else if (pre[str[j + 1]] != str[j]) flag = false;
            }
            v[str[len - 1]] = true;
        }
        for (int i = 'a'; i <= 'z'; i++) {
            memset(g, false, sizeof(g));
            if (!dfs(i)) flag = false;
        }
        if (!flag) printf("NO
");
        else {
            len = 0;
            for (int i = 'a'; i <= 'z'; i++) {
                if (v[i]) {
                    bool root = true;
                    for (int j = 'a'; j <= 'z'; j++) if (next[j] == i) root = false;
                    if (!root) continue;
                    int ptr = i;
                    while (ptr != 0) {
                        str[len++] = ptr;
                        v[ptr] = false;
                        ptr = next[ptr];
                    }
                }
            }
            str[len] = '';
            printf("%s
", str);
        }
    }
    return 0;
}

Maximum Element

You asked to find the number of permutations p of length n such that exists index i, such that pi ≠ n, pi is greater than any pj for j in [1, i - 1] and greater then any pj for j in [i + 1, i + k]. We will call such permutations good.

Define D(n) as number of good permutations that have pn = n. Notice that if k ≥ n, then D(n) = 0. Let w be a permutations such that wn = n. If index of element n - 1 is lesser than n - k, then w is good. Otherwise if n - 1 index is j, j ≥ n - k, then because there are less then k elements between n - 1 and n, w could be good only if i from the definition would be lesser than j. In that case permutation w1, ..., wj would form a good permutation of length j of some numbers with wj being the maximum.

Therefore the following equation is correct:

Which can be computed in O(n2), or in O(n) rewritten in the form

and using prefix sums for values .

The answer is than calculated as follows:

Complexity: O(n).

Solution

Symmetric Projections

原点到点p的投影点的长度可以表示为p的横纵坐标的线性组合，所以，如果点集对某一条直线投影后关于某个点中心对称，那么这个点是点集质心对这条直线的投影。

首先求出点集的质心，对于落在质心上的一个点或关于质心对称的一对点不影响结果，可以去掉。

若此时点集中点的数量为0，则有无数条直线满足题意。

否则，对于点集中第一个点P0，枚举投影后与其对称的点，最多n个，并验证相应直线是否符合题意。

Solution（这个代码可能会由于long long数据范围问题被cha，不过数据太弱，还是a了，就没再改了）

Mod Mod Mod

题意无法理解