UVA 10497 - Sweet Child Makes Trouble 高精度DP

Children are always sweet but they can sometimes make you feel bitter. In this problem, you will see
how Tintin, a five year’s old boy, creates trouble for his parents. Tintin is a joyful boy and is always
busy in doing something. But what he does is not always pleasant for his parents. He likes most to play
with household things like his father’s wristwatch or his mother’s comb. After his playing he places it
in some other place. Tintin is very intelligent and a boy with a very sharp memory. To make things
worse for his parents, he never returns the things he has taken for playing to their original places.
Think about a morning when Tintin has managed to ‘steal’ three household objects. Now, in how
many ways he can place those things such that nothing is placed in their original place. Tintin does not
like to give his parents that much trouble. So, he does not leave anything in a completely new place;
he merely permutes the objects.
Input
There will be several test cases. Each will have a positive integer less than or equal to 800 indicating
the number of things Tintin has taken for playing. Each integer will be in a line by itself. The input
is terminated by a ‘-1’ (minus one) in a single line, which should not be processed.
Output
For each test case print an integer indicating in how many ways Tintin can rearrange the things he has
taken.
Sample Input
2
3
4
-1
Sample Output
1
2
9

题意：一个小孩，趁家长不在，拿家里的n个 家具玩，玩了之后放回，而且好坏，一定不是原来的位置（每个都不是），问你有多少种放法

题解：设dp[i]表示  放回i个的方法数，那么  dp[i] = (i-1)*(dp[i-1]+dp[i-2]);

对于第i个数，放在序列的最后一个位置，它的位置一定是正确的，所以一定要和前面i−1个的其中一个交换位置才可以,那么如果选中位置上的物品为错误归放的，即为dp[i−1],如果选中的位置上的物品为正确归放的，即为dp[i−2]

//meek///#include<bits/stdc++.h>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include<iostream>
#include<bitset>
#include<vector>
#include <queue>
#include <map>
#include <set>
#include <stack>
using namespace std ;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define fi first
#define se second
#define MP make_pair
typedef long long ll;
 
const int N = +;
const int M = ;
const int inf = 0x3f3f3f3f;
const ll MOD = ;
 
#define MAX_L 20005 //最大长度，可以修改
 
class bign
{
public:
    int len, s[MAX_L];//数的长度，记录数组
//构造函数
    bign();
    bign(const char*);
    bign(int);
    bool sign;//符号 1正数 0负数
    string toStr() const;//转化为字符串，主要是便于输出
    friend istream& operator>>(istream &,bign &);//重载输入流
    friend ostream& operator<<(ostream &,bign &);//重载输出流
//重载复制
    bign operator=(const char*);
    bign operator=(int);
    bign operator=(const string);
//重载各种比较
    bool operator>(const bign &) const;
    bool operator>=(const bign &) const;
    bool operator<(const bign &) const;
    bool operator<=(const bign &) const;
    bool operator==(const bign &) const;
    bool operator!=(const bign &) const;
//重载四则运算
    bign operator+(const bign &) const;
    bign operator++();
    bign operator++(int);
    bign operator+=(const bign&);
    bign operator-(const bign &) const;
    bign operator--();
    bign operator--(int);
    bign operator-=(const bign&);
    bign operator*(const bign &)const;
    bign operator*(const int num)const;
    bign operator*=(const bign&);
    bign operator/(const bign&)const;
    bign operator/=(const bign&);
//四则运算的衍生运算
    bign operator%(const bign&)const;//取模（余数）
    bign factorial()const;//阶乘
    bign Sqrt()const;//整数开根（向下取整）
    bign pow(const bign&)const;//次方
//一些乱乱的函数
    void clean();
    ~bign();
};
#define max(a,b) a>b ? a : b
#define min(a,b) a<b ? a : b
 
bign::bign()
{
    memset(s, , sizeof(s));
    len = ;
    sign = ;
}
 
bign::bign(const char *num)
{
    *this = num;
}
 
bign::bign(int num)
{
    *this = num;
}
 
string bign::toStr() const
{
    string res;
    res = "";
    for (int i = ; i < len; i++)
        res = (char)(s[i] + '') + res;
    if (res == "")
        res = "";
    if (!sign&&res != "")
        res = "-" + res;
    return res;
}
 
istream &operator>>(istream &in, bign &num)
{
    string str;
    in>>str;
    num=str;
    return in;
}
 
ostream &operator<<(ostream &out, bign &num)
{
    out<<num.toStr();
    return out;
}
 
bign bign::operator=(const char *num)
{
    memset(s, , sizeof(s));
    char a[MAX_L] = "";
    if (num[] != '-')
        strcpy(a, num);
    else
        for (int i = ; i < strlen(num); i++)
            a[i - ] = num[i];
    sign = !(num[] == '-');
    len = strlen(a);
    for (int i = ; i < strlen(a); i++)
        s[i] = a[len - i - ] - ;
    return *this;
}
 
bign bign::operator=(int num)
{
    char temp[MAX_L];
    sprintf(temp, "%d", num);
    *this = temp;
    return *this;
}
 
bign bign::operator=(const string num)
{
    const char *tmp;
    tmp = num.c_str();
    *this = tmp;
    return *this;
}
 
bool bign::operator<(const bign &num) const
{
    if (sign^num.sign)
        return num.sign;
    if (len != num.len)
        return len < num.len;
    for (int i = len - ; i >= ; i--)
        if (s[i] != num.s[i])
            return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i]));
    return !sign;
}
 
bool bign::operator>(const bign&num)const
{
    return num < *this;
}
 
bool bign::operator<=(const bign&num)const
{
    return !(*this>num);
}
 
bool bign::operator>=(const bign&num)const
{
    return !(*this<num);
}
 
bool bign::operator!=(const bign&num)const
{
    return *this > num || *this < num;
}
 
bool bign::operator==(const bign&num)const
{
    return !(num != *this);
}
 
bign bign::operator+(const bign &num) const
{
    if (sign^num.sign)
    {
        bign tmp = sign ? num : *this;
        tmp.sign = ;
        return sign ? *this - tmp : num - tmp;
    }
    bign result;
    result.len = ;
    int temp = ;
    for (int i = ; temp || i < (max(len, num.len)); i++)
    {
        int t = s[i] + num.s[i] + temp;
        result.s[result.len++] = t % ;
        temp = t / ;
    }
    result.sign = sign;
    return result;
}
 
bign bign::operator++()
{
    *this = *this + ;
    return *this;
}
 
bign bign::operator++(int)
{
    bign old = *this;
    ++(*this);
    return old;
}
 
bign bign::operator+=(const bign &num)
{
    *this = *this + num;
    return *this;
}
 
bign bign::operator-(const bign &num) const
{
    bign b=num,a=*this;
    if (!num.sign && !sign)
    {
        b.sign=;
        a.sign=;
        return b-a;
    }
    if (!b.sign)
    {
        b.sign=;
        return a+b;
    }
    if (!a.sign)
    {
        a.sign=;
        b=bign()-(a+b);
        return b;
    }
    if (a<b)
    {
        bign c=(b-a);
        c.sign=false;
        return c;
    }
    bign result;
    result.len = ;
    for (int i = , g = ; i < a.len; i++)
    {
        int x = a.s[i] - g;
        if (i < b.len) x -= b.s[i];
        if (x >= ) g = ;
        else
        {
            g = ;
            x += ;
        }
        result.s[result.len++] = x;
    }
    result.clean();
    return result;
}
 
bign bign::operator * (const bign &num)const
{
    bign result;
    result.len = len + num.len;
 
    for (int i = ; i < len; i++)
        for (int j = ; j < num.len; j++)
            result.s[i + j] += s[i] * num.s[j];
 
    for (int i = ; i < result.len; i++)
    {
        result.s[i + ] += result.s[i] / ;
        result.s[i] %= ;
    }
    result.clean();
    result.sign = !(sign^num.sign);
    return result;
}
 
bign bign::operator*(const int num)const
{
    bign x = num;
    bign z = *this;
    return x*z;
}
bign bign::operator*=(const bign&num)
{
    *this = *this * num;
    return *this;
}
 
bign bign::operator /(const bign&num)const
{
    bign ans;
    ans.len = len - num.len + ;
    if (ans.len < )
    {
        ans.len = ;
        return ans;
    }
 
    bign divisor = *this, divid = num;
    divisor.sign = divid.sign = ;
    int k = ans.len - ;
    int j = len - ;
    while (k >= )
    {
        while (divisor.s[j] == ) j--;
        if (k > j) k = j;
        char z[MAX_L];
        memset(z, , sizeof(z));
        for (int i = j; i >= k; i--)
            z[j - i] = divisor.s[i] + '';
        bign dividend = z;
        if (dividend < divid) { k--; continue; }
        int key = ;
        while (divid*key <= dividend) key++;
        key--;
        ans.s[k] = key;
        bign temp = divid*key;
        for (int i = ; i < k; i++)
            temp = temp * ;
        divisor = divisor - temp;
        k--;
    }
    ans.clean();
    ans.sign = !(sign^num.sign);
    return ans;
}
 
bign bign::operator/=(const bign&num)
{
    *this = *this / num;
    return *this;
}
 
bign bign::operator%(const bign& num)const
{
    bign a = *this, b = num;
    a.sign = b.sign = ;
    bign result, temp = a / b*b;
    result = a - temp;
    result.sign = sign;
    return result;
}
 
bign bign::pow(const bign& num)const
{
    bign result = ;
    for (bign i = ; i < num; i++)
        result = result*(*this);
    return result;
}
 
bign bign::factorial()const
{
    bign result = ;
    for (bign i = ; i <= *this; i++)
        result *= i;
    return result;
}
 
void bign::clean()
{
    if (len == ) len++;
    while (len >  && s[len - ] == '\0')
        len--;
}
 
bign bign::Sqrt()const
{
    if(*this<)return -;
    if(*this<=)return *this;
    bign l=,r=*this,mid;
    while(r-l>)
    {
        mid=(l+r)/;
        if(mid*mid>*this)
            r=mid;
        else
            l=mid;
    }
    return l;
}
 
bign::~bign()
{
}
 
bign dp[N];
void init() {
    dp[] = ;
    dp[] = ;
    bign tmp = ;
    for(int i=;i<=;i=i+) {
        dp[i] = (tmp)*(dp[i-] + dp[i-]);
    tmp+=;
    }
}
int main() {
    init();
    int n;
    while(scanf("%d",&n)!=EOF) {
        if(n==-) break;
       cout<<dp[n]<<endl;
    }
    return ;
}

代码