2015 HNU warm up 02

D - Four-Tower Towers of Hanoi

假设在4个柱子上挪n个个盘子的最小步骤是F[n]，在三个柱子上挪n个盘子的最小步骤是T[n]，则四柱汉诺塔问题，可以分成两部分考虑：

1.先把上面k个盘子用F[k]步挪到B盘；

2.把下面n-k个盘子用T[n-k]步挪到D盘；

3.把B盘上的k个盘子再用F[k]步挪到D盘。

所以可以得到递推式 F[n] = min{ 2 * F[k] + T[n-k] }。

T[n]的公式已经推出：2^n - 1。

题目还有一个要注意的地方：虽然题目保证最后答案不会超过int64的范围，但是计算过程却可能会。

//code source from sk#include<cstdio>#include<cstdlib>#include<iostream>#include<cstring>using namespace std;const int maxn = 1005;const unsigned __int64 INF = ((__int64)0 - 1);unsigned __int64 dp[maxn];unsigned __int64 two[maxn];void init(){    two[0] = 1;    for(int i = 1; i < maxn; i++)        two[i] = two[i-1]*2;    dp[0] = 0;    dp[1] = 1;    dp[2] = 3;    dp[3] = 5;    dp[4] = 9;    dp[5] = 13;    for(int i = 6; i < maxn; i++)    {        unsigned __int64 p = INF;        for(int k = i-1; k < i && i-k < 62; k--)            p = min(p, dp[k]*2+two[i-k]-1);        dp[i] = p;    }//    for(int i = 1; i < maxn; i++)//        cout << "dp[" << i << "]:" << dp[i] << endl;}int main(){    init();    int n;    int kase = 1;    while(scanf("%d", &n)!=EOF)    {        printf("Case %d: %I64d\n", kase++, dp[n]);    }    return 0;}

E - Light Switches

大素数判定 + dfs所有约数

代码许多地方参考了小伙伴的源码还有shoutmon学长的模板。

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <stdio.h>#include <math.h>#include <time.h>using namespace std;#define LL __int64#define INF 99999999999999LL#define Rand_time 5#define C 16381bool cmp(LL a, LL b){    return a < b;}LL MIN;LL gcd(LL a, LL b){    return a % b ? gcd(b, a % b) : b;}//(a * b) % nLL Multi_mod(LL a, LL b, LL n){    LL s = 0;    while(b)    {        if(b & 1) s = (s + a) % n;        a = (a + a) % n;        b >>= 1;    }    return s;}//(a ^ b) % nLL Pow_mod(LL a, LL b, LL n){    LL s = 1;    while(b)    {        if(b & 1) s = Multi_mod(s, a, n);        a = Multi_mod(a, a, n);        b >>= 1;    }    return s;}//默认1为合数bool Miller_Rabin(LL n){    LL u = n - 1, pre, x;    if(n == 2 || n == 3 || n == 5 || n == 7 || n == 11) return true;    if(n == 1 || !(n % 2) || !(n % 3) || !(n % 5) || !(n % 7) || !(n % 11)) return false;    //把 n-1 分解成 k 个 2 和 奇数 v 的乘积    //则可以把计算 a ^ (n-1) % n，分解成计算 k 次 (a^2) 迭代计算和 a^v % n 的计算    int k = 0;    while(!(u & 1)) u >>= 1, k++;    srand((LL)time(NULL));    for(int i = 0; i < Rand_time; i++)    {        x = rand() % (n - 2) + 2;        x = Pow_mod(x, u, n);        pre = x; //保留x,既x^2 mod n =1 的根        for(int j = 0; j < k; j++)        {            x = Multi_mod(x, x, n);            //如果n是素数那么 x^2mod n = 1，仅有两个根（不同余），+1和-1（n-1）            if(x == 1 && pre != 1 && pre != (n - 1))                return false;            pre = x;        }        if(x != 1) return false;    }    return true;}//在O(sqrt(p))的时间复杂度内找到n的一个小因子pLL Pollard_rho(LL n, int c){    LL x, y, d;    int i = 1, j = 2;    srand((LL)time(NULL));    x = rand() % (n - 1) + 1;    y = x;    while(1)    {        i++;        x = (Multi_mod(x, x, n) + c) % n;        d = gcd(y - x + n, n);        if(d != 1 && d != n) return d;        if(x == y) return n;        if(i == j)        {            y = x;            j <<= 1;        }    }}//求所有素因子LL fac[100];int cnt[100], k, ck;void find_factor(LL n, int c){    LL r, d;    if(n <= 1) return ;    if(Miller_Rabin(n))    {        fac[k++] = n;        return ;    }    r = Pollard_rho(n, c--);    d = n / r;    find_factor(d, c);    find_factor(r, c);}LL n, t, b;void deal(){    k = ck = 0;    find_factor(b, C);    sort(fac, fac + k, cmp);    cnt[0] = 1;    for(int i = 1; i < k; i++)    {        if(fac[i] == fac[ck]) cnt[ck]++;        else { fac[++ck] = fac[i]; cnt[ck] = 1; }    }    ck++;}LL yinzi[12345];int num;void dfs(LL val, int k){    if(k == ck) { yinzi[num++] = val; return ;}    for(int i = 0, v = 1; i <= cnt[k]; i++)    {        dfs(val * v, k + 1);        v *= fac[k];    }}char res[2][4] = {"On", "Off"};int cas = 0;void prt(int w){    printf("Case %d: %s\n", ++cas, res[w]);}int main(){//    freopen("Light_Switches.in", "r", stdin);    while(~scanf("%I64d%I64d%I64d", &n, &t, &b))    {        t %= n; if(!t) t = n;        if(b == 1LL) { prt(0); continue; }        if(Miller_Rabin(b))        {            if(t >= b) prt(1);            else prt(0);            continue;        }        deal();        num = 0;        dfs(1, 0);        sort(yinzi, yinzi + num, cmp);//        for(int i = 0; i < num; i++) cout << yinzi[i] << ' '; cout << endl;        int id = lower_bound(yinzi, yinzi + num, t) - yinzi;        prt((id + 1) & 1);    }    return 0;}