Regionals 2012 :: Chengdu

题目连接

排行榜

A和I都是签到题

按位BFS K Yet Another Multiple Problem

题意：给一些可以用的数字，求最小的数，它由特定的数字组成且是n的倍数

分析：暴力枚举不可行，因为数字可能非常大。考虑到大数取模为0，BFS每一层即数位，递归输出路径。

#include <bits/stdc++.h>

const int N = 1e4 + 5;
bool no[10];
std::pair<int, int> pre[N];
int dir[10];
bool vis[N];
int n, m, tot;

void print(int u) {
    if (u != 0) {
        print (pre[u].first);
        printf ("%d", pre[u].second);
    }
    return ;
}

void BFS() {
    memset (vis, false, sizeof (vis));
    std::queue<int> que;
    que.push (0);
    while (!que.empty ()) {
        int x = que.front (); que.pop ();
        for (int i=0; i<tot; ++i) {
            if (x == 0 && dir[i] == 0) {
                continue;
            }
            int y = (x * 10 + dir[i]) % n;
            if (y == 0) {
                print (x);
                printf ("%d\n", dir[i]);
                return ;
            }
            if (vis[y]) {
                continue;
            }
            vis[y] = true;
            pre[y] = std::make_pair (x, dir[i]);
            que.push (y);
        }
    }
    puts ("-1");
    return ;
}

int main() {
    int cas = 0;
    while (scanf ("%d%d", &n, &m) == 2) {
        memset (no, false, sizeof (no));
        for (int i=0; i<m; ++i) {
            int x; scanf ("%d", &x);
            no[x] = true;
        }
        tot = 0;
        for (int i=0; i<10; ++i) {
            if (!no[i]) {
                dir[tot++] = i;
            }
        }
        printf ("Case %d: ", ++cas);
        BFS ();
    }

    return 0;
}

同类型的题目：

URAL 1495

#include <bits/stdc++.h>

const int N = 1e6 + 5;
std::pair<int, int> pre[N];
bool vis[N];
int n;

void print(int u) {
    if (u != 0) {
        print (pre[u].first);
        printf ("%d", pre[u].second);
    }
    return ;
}

void BFS() {
    memset (vis, false, sizeof (vis));
    std::queue<int> que;
    que.push (0);
    while (!que.empty ()) {
        int x = que.front (); que.pop ();
        for (int i=1; i<3; ++i) {
            int y = (x * 10 + i) % n;
            if (y == 0) {
                print (x);
                printf ("%d\n", i);
                return ;
            }
            if (vis[y]) {
                continue;
            }
            vis[y] = true;
            pre[y] = std::make_pair (x, i);
            que.push (y);
        }
    }
    puts ("Impossible");
    return ;
}

int main() {
    while (scanf ("%d", &n) == 1) {
        BFS ();
    }

    return 0;
}

HDOJ 1226

#include <bits/stdc++.h>

const int N = 5e3 + 5;
std::pair<int, int> pre[N];
struct Node {
    int x, len;
};
int dir[16];
bool vis[N];
int n, c, m;

void print(int u) {
    if (u != 0) {
        print (pre[u].first);
        int x = pre[u].second;
        if (x < 10) {
            printf ("%d", x);
        } else {
            printf ("%c", 'A' + x - 10);
        }
    }
    return ;
}

void BFS() {
    memset (vis, false, sizeof (vis));
    std::queue<Node> que;
    que.push ((Node) {0, 0});
    while (!que.empty ()) {
        Node &u = que.front (); que.pop ();
        if (u.len >= 500) {
            continue;
        }
        for (int i=0; i<m; ++i) {
            if (u.x == 0 && dir[i] == 0) {
                continue;
            }
            int y = (u.x * c + dir[i]) % n;
            if (y == 0) {
                print (u.x);
                int d = dir[i];
                if (d < 10) {
                    printf ("%d\n", d);
                } else {
                    printf ("%c\n", 'A' + d - 10);
                }
                return ;
            }
            if (vis[y]) {
                continue;
            }
            vis[y] = true;
            pre[y] = std::make_pair (u.x, dir[i]);
            que.push ((Node) {y, u.len + 1});
        }
    }
    puts ("give me the bomb please");
    return ;
}

int main() {
    int T; scanf ("%d", &T);
    while (T--) {
        scanf ("%d%d", &n, &c);
        scanf ("%d", &m);
        char str[3];
        for (int i=0; i<m; ++i) {
            scanf ("%s", str);
            if (str[0] >= 'A' && str[0] <= 'F') {
                dir[i] = str[0] - 'A' + 10;
            } else {
                dir[i] = str[0] - '0';
            }
        }
        std::sort (dir, dir+m);
        if (n == 0) {
            if (dir[0] == 0) {
                puts ("0");
            } else {
                puts ("give me the bomb please");
            }
        } else {
            BFS ();
        }
    }

    return 0;
}

数学期望 B Candy

题意：两堆n个物品，每次拿走一个，从第一堆拿的概率p，另一堆概率1-p，问其中一堆0时，另一堆数量的期望。

分析：公式为：。p^n不好直接算，技巧：取对数后在exp阶乘回来

#include <bits/stdc++.h>

//ret = sigma(exp(log(C(n+i, i) + (n+1) * log(p) + i * log(1-p))));
double run(int n, double p) {
    double ret = 0, lp = log (p), lq = log (1-p), c = log (1.0);
    ret = exp (c + (n+1) * lp + 0 * lq) * n;
    for (int i=1; i<n; ++i) {
        c = c + log (n + i) - log (i);
        ret += exp (c + (n+1) * lp + i * lq) * (n - i);
    }
    return ret;
}

int main() {
    int n, cas = 0; double p;
    while (scanf ("%d%lf", &n, &p) == 2) {
        printf ("Case %d: ", ++cas);
        if (p == 0 || p == 1) {
            printf ("%.8f\n", (double) n);
        } else {
            printf ("%.8f\n", run (n, p) + run (n, 1.0 - p));
        }
    }
    return 0;
}

数论 J Exam

题意：定义f(x) = 满足x%(a*b)=0的pair(a,b)的数量，求f(1)+f(2)+f(3)+...+f(n)

分析：n<=10^11普通枚举不可行。转换一下，问题变成x=a*b*c的pair(a,b,c)的数量，设a<=b<=c，则a<=,b<=，枚举a和b，复杂度为O().还一个问题，题目求前缀总和，考虑pair(a,b)对1~n的贡献为n/(a*b)，相当于1~n是a*b的倍数的个数。

#include <bits/stdc++.h>

typedef long long ll;

int sqrt2(ll x) {
    ll ret = (int) pow (1.0 * x, 0.5);
    while (ret * ret < x) {
        ret++;
    }
    while (ret * ret > x) {
        ret--;
    }
    return ret;
}

int sqrt3(ll x) {
    ll ret = (int) pow (1.0 * x, 1.0 / 3);
    while (ret * ret * ret < x) {
        ret++;
    }
    while (ret * ret * ret > x) {
        ret--;
    }
    return ret;
}

ll run(ll n) {
    ll sq3 = sqrt3 (n);
    ll ret = sq3; //a = b = c
    for (int i=1; i<=sq3; ++i) {
        ll ni = n / i;
        ll k = sqrt2 (ni);
        ret += (ni / i - i) * 3; //a = b < c
        ret += (k - i) * 3;     //a < b = c
        for (int j=i+1; j<=k; ++j) {
            ret += (ni / j - j) * 6; //a < b < c
        }
    }
    return ret;
}

int main() {
    ll n;
    int cas = 0;
    while (scanf ("%I64d", &n) == 1) {
        printf ("Case %d: %I64d\n", ++cas, run (n));
    }
    return 0;
}