UVA 11375 Matches

UVA 11375  Matches

题意：用火柴棍搭数字，能搭多少个非负整数(不能有前导0，数字0是可以的，6根火柴即可搭出)。设一共有n根火柴。

思路：dp+大整数 设dp[i]:用i根火柴正好能搭的数字的个数  设搭出每个数字j(0<=j<=9)需要的火柴数为c[j].

那么动态转移方程：dp[i+c[j]]+=dp[i];

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<string>
using namespace std;
const int N_MAX = 2000 + 20;
#define MAX_L 2005 //最大长度，可以修改  
typedef unsigned long long ll;

class bign
{
public:
    int len, s[MAX_L];//数的长度，记录数组  
                      //构造函数  
    bign();
    bign(const char*);
    bign(int);
    bool sign;//符号 1正数 0负数  
    string toStr() const;//转化为字符串，主要是便于输出  
    friend istream& operator >> (istream &, bign &);//重载输入流  
    friend ostream& operator<<(ostream &, bign &);//重载输出流  
                                                  //重载复制  
    bign operator=(const char*);
    bign operator=(int);
    bign operator=(const string);
    //重载各种比较  
    bool operator>(const bign &) const;
    bool operator>=(const bign &) const;
    bool operator<(const bign &) const;
    bool operator<=(const bign &) const;
    bool operator==(const bign &) const;
    bool operator!=(const bign &) const;
    //重载四则运算  
    bign operator+(const bign &) const;
    bign operator++();
    bign operator++(int);
    bign operator+=(const bign&);
    bign operator-(const bign &) const;
    bign operator--();
    bign operator--(int);
    bign operator-=(const bign&);
    bign operator*(const bign &)const;
    bign operator*(const int num)const;
    bign operator*=(const bign&);
    bign operator/(const bign&)const;
    bign operator/=(const bign&);
    //四则运算的衍生运算  
    bign operator%(const bign&)const;//取模（余数）  
    bign factorial()const;//阶乘  
    bign Sqrt()const;//整数开根（向下取整）  
    bign pow(const bign&)const;//次方  
                               //一些乱乱的函数  
    void clean();
    ~bign();
};
#define max(a,b) a>b ? a : b  
#define min(a,b) a<b ? a : b  

bign::bign()
{
    memset(s, 0, sizeof(s));
    len = 1;
    sign = 1;
}

bign::bign(const char *num)
{
    *this = num;
}

bign::bign(int num)
{
    *this = num;
}

string bign::toStr() const
{
    string res;
    res = "";
    for (int i = 0; i < len; i++)
        res = (char)(s[i] + '0') + res;
    if (res == "")
        res = "0";
    if (!sign&&res != "0")
        res = "-" + res;
    return res;
}

istream &operator >> (istream &in, bign &num)
{
    string str;
    in >> str;
    num = str;
    return in;
}

ostream &operator<<(ostream &out, bign &num)
{
    out << num.toStr();
    return out;
}

bign bign::operator=(const char *num)
{
    memset(s, 0, sizeof(s));
    char a[MAX_L] = "";
    if (num[0] != '-')
        strcpy(a, num);
    else
        for (int i = 1; i < strlen(num); i++)
            a[i - 1] = num[i];
    sign = !(num[0] == '-');
    len = strlen(a);
    for (int i = 0; i < strlen(a); i++)
        s[i] = a[len - i - 1] - 48;
    return *this;
}

bign bign::operator=(int num)
{
    if (num < 0)
        sign = 0, num = -num;
    else
        sign = 1;
    char temp[MAX_L];
    sprintf(temp, "%d", num);
    *this = temp;
    return *this;
}

bign bign::operator=(const string num)
{
    const char *tmp;
    tmp = num.c_str();
    *this = tmp;
    return *this;
}

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

bool bign::operator>(const bign&num)const
{
    return num < *this;
}

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

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

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

bool bign::operator==(const bign&num)const
{
    return !(num != *this);
}

bign bign::operator+(const bign &num) const
{
    if (sign^num.sign)
    {
        bign tmp = sign ? num : *this;
        tmp.sign = 1;
        return sign ? *this - tmp : num - tmp;
    }
    bign result;
    result.len = 0;
    int temp = 0;
    for (int i = 0; temp || i < (max(len, num.len)); i++)
    {
        int t = s[i] + num.s[i] + temp;
        result.s[result.len++] = t % 10;
        temp = t / 10;
    }
    result.sign = sign;
    return result;
}

bign bign::operator++()
{
    *this = *this + 1;
    return *this;
}

bign bign::operator++(int)
{
    bign old = *this;
    ++(*this);
    return old;
}

bign bign::operator+=(const bign &num)
{
    *this = *this + num;
    return *this;
}

bign bign::operator-(const bign &num) const
{
    bign b = num, a = *this;
    if (!num.sign && !sign)
    {
        b.sign = 1;
        a.sign = 1;
        return b - a;
    }
    if (!b.sign)
    {
        b.sign = 1;
        return a + b;
    }
    if (!a.sign)
    {
        a.sign = 1;
        b = bign(0) - (a + b);
        return b;
    }
    if (a<b)
    {
        bign c = (b - a);
        c.sign = false;
        return c;
    }
    bign result;
    result.len = 0;
    for (int i = 0, g = 0; i < a.len; i++)
    {
        int x = a.s[i] - g;
        if (i < b.len) x -= b.s[i];
        if (x >= 0) g = 0;
        else
        {
            g = 1;
            x += 10;
        }
        result.s[result.len++] = x;
    }
    result.clean();
    return result;
}

bign bign::operator * (const bign &num)const
{
    bign result;
    result.len = len + num.len;

    for (int i = 0; i < len; i++)
        for (int j = 0; j < num.len; j++)
            result.s[i + j] += s[i] * num.s[j];

    for (int i = 0; i < result.len; i++)
    {
        result.s[i + 1] += result.s[i] / 10;
        result.s[i] %= 10;
    }
    result.clean();
    result.sign = !(sign^num.sign);
    return result;
}

bign bign::operator*(const int num)const
{
    bign x = num;
    bign z = *this;
    return x*z;
}
bign bign::operator*=(const bign&num)
{
    *this = *this * num;
    return *this;
}

bign bign::operator /(const bign&num)const
{
    bign ans;
    ans.len = len - num.len + 1;
    if (ans.len < 0)
    {
        ans.len = 1;
        return ans;
    }

    bign divisor = *this, divid = num;
    divisor.sign = divid.sign = 1;
    int k = ans.len - 1;
    int j = len - 1;
    while (k >= 0)
    {
        while (divisor.s[j] == 0) j--;
        if (k > j) k = j;
        char z[MAX_L];
        memset(z, 0, sizeof(z));
        for (int i = j; i >= k; i--)
            z[j - i] = divisor.s[i] + '0';
        bign dividend = z;
        if (dividend < divid) { k--; continue; }
        int key = 0;
        while (divid*key <= dividend) key++;
        key--;
        ans.s[k] = key;
        bign temp = divid*key;
        for (int i = 0; i < k; i++)
            temp = temp * 10;
        divisor = divisor - temp;
        k--;
    }
    ans.clean();
    ans.sign = !(sign^num.sign);
    return ans;
}

bign bign::operator/=(const bign&num)
{
    *this = *this / num;
    return *this;
}

bign bign::operator%(const bign& num)const
{
    bign a = *this, b = num;
    a.sign = b.sign = 1;
    bign result, temp = a / b*b;
    result = a - temp;
    result.sign = sign;
    return result;
}

bign bign::pow(const bign& num)const
{
    bign result = 1;
    for (bign i = 0; i < num; i++)
        result = result*(*this);
    return result;
}

bign bign::factorial()const
{
    bign result = 1;
    for (bign i = 1; i <= *this; i++)
        result *= i;
    return result;
}

void bign::clean()
{
    if (len == 0) len++;
    while (len > 1 && s[len - 1] == '')
        len--;
}

bign bign::Sqrt()const
{
    if (*this<0)return -1;
    if (*this <= 1)return *this;
    bign l = 0, r = *this, mid;
    while (r - l>1)
    {
        mid = (l + r) / 2;
        if (mid*mid>*this)
            r = mid;
        else
            l = mid;
    }
    return l;
}

bign::~bign()
{
}


int c[10] = { 6,2,5,5,4,5,6,3,7,6 };//每个数字i需要用多少火柴构成
bign dp[N_MAX];//用i根火柴能构成多少数字
int main() {
    memset(dp, 0, sizeof(dp));
        dp[0] = 1;
        for (int i = 0; i < N_MAX; i++) {
            for (int j = 0; j < 10; j++) {
                if (!(!i && !j) && i + c[j] < N_MAX) {
                    dp[i + c[j]] += dp[i];
                }
            }
        }
    int n;
    while (scanf("%d", &n) != EOF) {
        bign sum = 0;
        for (int i = 1; i <= n; i++) {
            sum += dp[i];
        }
        if (n >= 6)sum += 1;
        cout << sum << endl;
    }
    return 0;
}