UVA 10497 - Sweet Child Makes Trouble（DP+高精度）

题目链接：10497 - Sweet Child Makes Trouble

题意：n个物品，原来物品属于一个地方，现在要把物品重新放回去，问能放几种使得每个物品都与原来位置不同

思路：递推，一开始随便搞了个二维状态，dp[i][j]表示i个物品，有j个位置不同，那么dp[n][n]就是答案，递推式为：

dp[i][j] = 1 (j == 0)

dp[i][j] = (j - 1) * dp[i - 1][j - 1] + dp[i - 1][j] + (i - j + 1) * dp[i - 1][j - 2]; ( j > 1)

然后跑一下最大的n = 800，发现longlong都存不下，然后改高精度，发现加上高精度后这个状态是开不下的。

然后重新想状态，明显只够一维了，dp[i]表示i个物品的情况。

那么可以这样考虑，dp[i] 从 dp[i - 1]的状态过来，随便把一个位置和最后多上的那个位置对调就满足了，可行的位置一共有i - 1个，种数为(i - 1) * dp[i - 1];

dp[i]从dp[i - 2]状态过来，这样有两个相同的，把这两个位置交换一下也满足了，这位置一共也有i - 1个，种数为(i - 1) * dp[i - 2]；

于是状态转移出来了，dp[i] = (i - 1) * ( dp[i - 1] + dp[i - 2]);

代码：

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;
const int MAXN = 2005;

struct bign {
	int len, num[MAXN];

	bign () {
		len = 0;
		memset(num, 0, sizeof(num));
	}
	bign (int number) {*this = number;}
	bign (const char* number) {*this = number;}

	void DelZero ();
	void Put ();

	void operator = (int number);
	void operator = (char* number);

	bool operator <  (const bign& b) const;
	bool operator >  (const bign& b) const { return b < *this; }
	bool operator <= (const bign& b) const { return !(b < *this); }
	bool operator >= (const bign& b) const { return !(*this < b); }
	bool operator != (const bign& b) const { return b < *this || *this < b;}
	bool operator == (const bign& b) const { return !(b != *this); }

	void operator ++ ();
	void operator -- ();
	bign operator + (const int& b);
	bign operator + (const bign& b);
	bign operator - (const int& b);
	bign operator - (const bign& b);
	bign operator * (const int& b);
	bign operator * (const bign& b);
	bign operator / (const int& b);
	//bign operator / (const bign& b);
	int operator % (const int& b);
};

/*Code*/
bign dp[805];

int main() {
	dp[0] = 0; dp[1] = 0; dp[2] = 1;
	for (int i = 3; i <= 800; i++) {
		dp[i] = (dp[i - 1] + dp[i - 2]) * (i - 1);
 	}
 	int n;
 	while (~scanf("%d", &n) && n >= 0) {
 		dp[n].Put();
 		printf("\n");
  	}
	return 0;
}
void bign::DelZero () {
	while (len && num[len-1] == 0)
		len--;

	if (len == 0) {
		num[len++] = 0;
	}
}

void bign::Put () {
	for (int i = len-1; i >= 0; i--)
		printf("%d", num[i]);
}

void bign::operator = (char* number) {
	len = strlen (number);
	for (int i = 0; i < len; i++)
		num[i] = number[len-i-1] - ‘0‘;

	DelZero ();
}

void bign::operator = (int number) {

	len = 0;
	while (number) {
		num[len++] = number%10;
		number /= 10;
	}

	DelZero ();
}

bool bign::operator < (const bign& b) const {
	if (len != b.len)
		return len < b.len;
	for (int i = len-1; i >= 0; i--)
		if (num[i] != b.num[i])
			return num[i] < b.num[i];
	return false;
}

void bign::operator ++ () {
	int s = 1;

	for (int i = 0; i < len; i++) {
		s = s + num[i];
		num[i] = s % 10;
		s /= 10;
		if (!s) break;
	}

	while (s) {
		num[len++] = s%10;
		s /= 10;
	}
}

void bign::operator -- () {
	if (num[0] == 0 && len == 1) return;

	int s = -1;
	for (int i = 0; i < len; i++) {
		s = s + num[i];
		num[i] = (s + 10) % 10;
		if (s >= 0) break;
	}
	DelZero ();
}

bign bign::operator + (const int& b) {
	bign a = b;
	return *this + a;
}

bign bign::operator + (const bign& b) {
	int bignSum = 0;
	bign ans;

	for (int i = 0; i < len || i < b.len; i++) {
		if (i < len) bignSum += num[i];
		if (i < b.len) bignSum += b.num[i];

		ans.num[ans.len++] = bignSum % 10;
		bignSum /= 10;
	}

	while (bignSum) {
		ans.num[ans.len++] = bignSum % 10;
		bignSum /= 10;
	}

	return ans;
}

bign bign::operator - (const int& b) {
	bign a = b;
	return *this - a;
}

bign bign::operator - (const bign& b) {
	int bignSub = 0;
	bign ans;
	for (int i = 0; i < len || i < b.len; i++) {
		bignSub += num[i];
		bignSub -= b.num[i];
		ans.num[ans.len++] = (bignSub + 10) % 10;
		if (bignSub < 0) bignSub = -1;
	}
	ans.DelZero ();
	return ans;
}

bign bign::operator * (const int& b) {
	int bignSum = 0;
	bign ans;

	ans.len = len;
	for (int i = 0; i < len; i++) {
		bignSum += num[i] * b;
		ans.num[i] = bignSum % 10;
		bignSum /= 10;
	}

	while (bignSum) {
		ans.num[ans.len++] = bignSum % 10;
		bignSum /= 10;
	}

	return ans;
}

bign bign::operator * (const bign& b) {
	bign ans;
	ans.len = 0; 

	for (int i = 0; i < len; i++){
		int bignSum = 0;  

		for (int j = 0; j < b.len; j++){
			bignSum += num[i] * b.num[j] + ans.num[i+j];
			ans.num[i+j] = bignSum % 10;
			bignSum /= 10;
		}
		ans.len = i + b.len;  

		while (bignSum){
			ans.num[ans.len++] = bignSum % 10;
			bignSum /= 10;
		}
	}
	return ans;
}

bign bign::operator / (const int& b) {

	bign ans;

	int s = 0;
	for (int i = len-1; i >= 0; i--) {
		s = s * 10 + num[i];
		ans.num[i] = s/b;
		s %= b;
	}

	ans.len = len;
	ans.DelZero ();
	return ans;
}

int bign::operator % (const int& b) {

	bign ans;

	int s = 0;
	for (int i = len-1; i >= 0; i--) {
		s = s * 10 + num[i];
		ans.num[i] = s/b;
		s %= b;
	}

	return s;
}

UVA 10497 - Sweet Child Makes Trouble（DP+高精度）,布布扣,bubuko.com