uva 10844 - Bloques(数论+高精度)

题目链接：uva 10844 - Bloques

题目大意：给出一个n，表示有1~n这n个数，问有多少种划分子集的方法。

解题思路：递推+高精度。

1
1 2
2 3 5
5 7 10 15
15 20 27 37 52

• dp[i][j]=dp[i−1][j−1]+dp[i][j−1]
• dp[i][0]=dp[i−1][i−1]
• ans[i]=dp[i][i]

需要用到高精度，并且缩进。

#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int MAXN = 1005;const int MOD = 1e8;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;bign dp[2][N], ans[N];void init () {    ans[1] = 1;    dp[1][1] = 1;    for (int i = 2; i < N; i++) {        int u = i%2;        int v = 1-u;        dp[u][1] = ans[i-1];;        for (int j = 2; j <= i; j++)            dp[u][j] = dp[u][j-1] + dp[v][j-1];        ans[i] = dp[u][i];    }}int main () {    int n;    init ();    while (scanf("%d", &n) == 1 && 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%MOD;        number /= MOD;    }    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 % MOD;        bignSum /= MOD;    }    while (bignSum) {        ans.num[ans.len++] = bignSum % MOD;        bignSum /= MOD;    }    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;}