题目链接:uva
10497 - Sweet Child Makes Trouble
题目大意:有个小屁孩很淘气,总是趁父母不在家的时候去拿家具玩,每次拿n个家具玩,但是放回去的时候一定不会将某个物品放回原处,一定要打乱。问说有多少放的方式满足。
解题思路:dp[i]表示说i个数打乱的情况,dp[i]=(dp[i?1]+dp[i?2])?(i?1)每次新增一个数,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]
#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) == 1 && 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)
时间: 2024-10-09 23:35:01