8VC Venture Cup 2016 - Elimination Round

 

在家补补题

 

模拟 A - Robot Sequence

#include <bits/stdc++.h>

char str[202];

void move(int &x, int &y, char ch)  {
    if (ch == 'U')  x--;
    if (ch == 'D')  x++;
    if (ch == 'L')  y--;
    if (ch == 'R')  y++;
}

int main(void)  {
    int n;  scanf ("%d", &n);
    scanf ("%s", &str);
    int ans = 0;
    for (int i=0; i<n; ++i) {
        int x = 0, y = 0;
        for (int j=i; j<n; ++j) {
            move (x, y, str[j]);
            if (x == 0 && y == 0)   ans++;
        }
    }
    printf ("%d\n", ans);

    return 0;
}

暴力 || 找规律 B - Cards

暴力即DFS也行。。。当时就if else乱写一堆过了

#include <bits/stdc++.h>

char str[202];
int col[3];

int main(void)  {
    int n;  scanf ("%d", &n);
    scanf ("%s", &str);
    col[0] = col[1] = col[2] = 0;
    for (int i=0; i<n; ++i) {
        if (str[i] == 'R')  col[0]++;
        if (str[i] == 'G')  col[1]++;
        if (str[i] == 'B')  col[2]++;
    }
    if (col[0] == n)    puts ("R");
    else if (col[1] == n) puts ("G");
    else if (col[2] == n)   puts ("B");
    else    {
        if (col[0] && col[1] && col[2]) puts ("BGR");
        else    {
            if (n == 2) {
                if (col[0] == 0)    puts ("R");
                else if (col[1] == 0)   puts ("G");
                else if (col[2] == 0)   puts ("B");
            }
            else if (col[0] == 1)   {
                if (col[1] == 0)    puts ("GR");
                if (col[2] == 0)    puts ("BR");
            }
            else if (col[1] == 1)   {
                if (col[0] == 0)    puts ("GR");
                if (col[2] == 0)    puts ("BG");
            }
            else if (col[2] == 1)    {
                if (col[0] == 0)    puts ("BR");
                if (col[1] == 0)    puts ("BG");
            }
            else    puts ("BGR");
        }
    }

    return 0;
} 

暴力 C - Block Towers

直接枚举答案，同时记录能整除2，整除2和3，只能整除3的个数，当遇到满足条件的就是最优。二分也行。。

#include <bits/stdc++.h>

const int N = 4e6 + 5;

int main(void)  {
    int n, m;   scanf ("%d%d", &n, &m);
    int best = 4000000;
    int c1 = 0, c2 = 0, c3 = 0;
    for (int i=2; i<=4000000; ++i) {
        if (i % 2 == 0) c1++;
        if (i % 2 == 0 && i % 3 == 0)   c2++;
        if (i % 2 != 0 && i % 3 == 0) c3++;
        if (c1 >= n && c2 >= m - c3 && c1 - (m - c3) >= n)  {
            best = i;   break;
        }
    }
    printf ("%d\n", best);

    return 0;
}

暴力+概率 D - Jerry's Protest

题意：已知结果两胜一负，问总和小于后者的概率。

分析：预处理出任意两个数字相减的差的方案数，那么前两次在正数选，后者只要能使得总和小的可以处理前缀和，O (1)，总复杂度 O (n ^ 2)。

#include <bits/stdc++.h>

const int N = 2e3 + 5;

int a[N];
int cnt[10005];
int sum[5005];

int main(void)  {
    int n;  scanf ("%d", &n);
    for (int i=1; i<=n; ++i)    scanf ("%d", &a[i]);
    for (int i=1; i<=n; ++i)    {
        for (int j=1; j<=n; ++j)    {
            if (i == j) continue;
            cnt[5000+a[i]-a[j]]++;
        }
    }
    for (int i=1; i<=5000; ++i) {
        sum[i] = sum[i-1] + cnt[i];
    }
    double ans = 0;
    for (int i=5001; i<=9999; ++i)  {
        if (!cnt[i])    continue;
        for (int j=5001; j<=9999; ++j)  {
            if (!cnt[j])    continue;
            int c = 15000 - (i + j) - 1;
            if (c < 1 || c > 4999 || !sum[c])  continue;
            ans += 1.0 * cnt[i] * cnt[j] * sum[c];
        }
    }
    double div = 1.0 * (n * (n - 1) / 2);
    ans = ans / div / div / div;
    printf ("%.8f\n", ans);

    return 0;
}

三分 + 贪心 E - Simple Skewness

题意：选取一个子集使得平均数-中位数最大

分析：首先个数是奇数，如果是偶数，去掉中间较大的数，差会变大(?)。然后排序后，枚举中位数的位置，三分长度，因为差的分布是单峰，选的数字使差尽可能大，右边选择最后几个，左边选取靠近中位数的几个。

#include <bits/stdc++.h>

typedef long long ll;
const int N = 2e5 + 5;
struct Pair {
    ll a;   int b;
    bool operator < (const Pair &rhs) const {
        return a * rhs.b < rhs.a * b;
    }
};
int a[N];
ll sum[N];
int n, bi, bl;
Pair best;

Pair get(int id, int len)   {
    ll val = sum[id] - sum[id-len-1];
    val += sum[n] - sum[n-len];
    Pair cur = Pair {val, 2 * len + 1};
    cur.a -= 1ll * a[id] * cur.b;
    if (best < cur) {
        best = cur;
        bi = id;    bl = len;
    }
    return cur;
}

int main(void)  {
    scanf ("%d", &n);
    for (int i=1; i<=n; ++i)    {
        scanf ("%d", a + i);
    }
    std::sort (a+1, a+1+n);
    for (int i=1; i<=n; ++i)    {
        sum[i] = sum[i-1] + a[i];
    }
    bi = 1; bl = 0;
    best = Pair {0, 1};
    for (int i=1; i<=n; ++i)    {
        int low = 0, high = std::min (i - 1, n - i);
        while (low + 3 < high)  {
            int mid1 = (2 * low + high) / 3;
            int mid2 = (low + 2 * high) / 3;
            Pair v1 = get (i, mid1);
            Pair v2 = get (i, mid2);
            if (v1 < v2)    low = mid1;
            else    high = mid2;
        }
        for (int j=low; j<=high; ++j)   get (i, j);
    }
    std::vector<int> ans;
    for (int i=bi-bl; i<=bi; ++i)   {
        ans.push_back (a[i]);
    }
    for (int i=n-bl+1; i<=n; ++i)  {
        ans.push_back (a[i]);
    }
    printf ("%d\n", ans.size ());
    for (int i=0; i<ans.size (); ++i)   {
        if (i > 0)  putchar (' ');
        printf ("%d", ans[i]);
    }
    puts ("");

    return 0;
}　

DP F - Group Projects

题意：n个数字分组，求每组最大值-最小值的和小于k的方案数

分析：明显的DP，复杂度肯定是 O (n ^ 2 * k)，难在状态的转移。dp[i][j][k] 考虑前i个数字，open(只知道最小值，最大值未知)了j组，当前累计和为k的方案数。那么open的几组暂时由a[i]"托管"，之前是a[i-1]“托管”，那么前后转移累加j * (a[i] - a[i-1])，a[i]有好几种选择，可以close一个组，选择一个组自己为最大值，j - 1；可以多一个组，自己为最小值，j+1；还可以进入某一个组(自己不是最大值)或者自己一个数字成为一个组(不算open)。

#include <bits/stdc++.h>

const int N = 2e2 + 5;
const int K = 1e3 + 5;
const int MOD = 1e9 + 7;
int a[N];
int dp[2][N][K];

void add(int &x, int y) {
    x += y;
    if (x >= MOD)   x %= MOD;
}

int main(void)  {
    int n, m;   scanf ("%d%d", &n, &m);
    for (int i=1; i<=n; ++i)    scanf ("%d", a + i);
    std::sort (a+1, a+1+n);
    int now = 0;
    dp[now][0][0] = 1;
    for (int i=1; i<=n; ++i)    {
        now ^= 1;
        memset (dp[now], 0, sizeof (dp[now]));
        for (int j=0; j<i; ++j) {
            for (int k=0; k<=m; ++k)    {
                if (!dp[now^1][j][k])    continue;
                int &x = dp[now^1][j][k];
                int s = k + j * (a[i] - a[i-1]);
                if (s > m)  continue;
                add (dp[now][j][s], 1ll * x * (j + 1) % MOD);
                add (dp[now][j+1][s], x);
                if (j)  add (dp[now][j-1][s], 1ll * x * j % MOD);
            }
        }
    }
    int ans = 0;
    for (int i=0; i<=m; ++i)    add (ans, dp[now][0][i]);
    printf ("%d\n", ans);

    return 0;
}