uva 10844 - Bloques(数论+高精度)

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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;
}

uva 10844 - Bloques(数论+高精度),布布扣,bubuko.com

uva 10844 - Bloques(数论+高精度)

标签：style   http   color   get   width   os   

原文地址：http://blog.csdn.net/keshuai19940722/article/details/35258329