UVA 10844 - Bloques

题目链接

题意：给定n个数字，问这n个数字能分成子集分成有几种分法

思路：一开始先想了个状态，dp[i][j]表示放i个数字，分成j个集合的方案，那么转移为，从dp[i - 1][j - 1]在多一个集合，和从dp[i - 1][j]有j个位置放，那么转移方程为dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] * j;按理说这个状态转移是没问题的，但是由于这题答案是高精度，n为900时答案高达1700多位，加上高精度运算结果就超时了，后面知道这种是bell数，是可以通过杨辉三角推出来的，具体就是一个数字等于杨辉三角的左边和左上边相加，这样状态转移变成纯高精度加法，然后把高精度加法运算进行压缩位的高精度运算，勉强通过了

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

using namespace std;
const int MAXN = 1800;

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*/

const int N = 905;
int n;
bign dp[2][N], ans[N];

void init() {
    int pre = 1, now = 0;
    dp[now][1] = 1; ans[1] = 1;
    for (int i = 2; i <= 900; i++) {
    swap(now, pre);
    dp[now][1] = dp[pre][i - 1];
    for (int j = 2; j <= i; j++)
        dp[now][j] = dp[now][j - 1] + dp[pre][j - 1];
    ans[i] = dp[now][i];
    }
}

int main() {
    init();
    while (~scanf("%d", &n) && n) {
    printf("%d, ", n);
    ans[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 () {
    printf("%d", num[len - 1]);
    for (int i = len-2; i >= 0; i--)
    printf("%08d", 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 % 100000000;
    bignSum /= 100000000;
    }

    while (bignSum) {
    ans.num[ans.len++] = bignSum % 100000000;
    bignSum /= 100000000;
    }

    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 10844 - Bloques (第二类斯特灵数)