UVa 11375 Matches

第一次用lrj的高精度类模板，感觉还是很好用的

c[x]表示数字x需要的火柴根数

将已经使用的火柴数i看做状态，每添加一个数字x状态就从i转移到i+c[x]

d[i]表示从节点0到节点i路径的条数，则答案f(n) = d(1) + d(2) + …… + d(n)

开始的时候不计入0，最后的时候如果n≥6答案再加一

#define LOCAL
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

struct bign
{
    int len, s[1000];
    bign() { memset(s, 0, sizeof(s)); len = 1; }
    bign operator = (const char* num)
    {
        len = strlen(num);
        for(int i = 0; i < len; ++i)
            s[i] = num[len - i - 1] - '0';
        return *this;
    }
    bign operator = (int num)
    {
        char s[1000];
        sprintf(s, "%d", num);
        *this = s;
        return *this;
    }
    bign (int num) { *this = num; }
    bign (const char* num) { *this = num; }
    string  str() const
    {
        string res = "";
        for(int i = 0; i < len; ++i)
            res = (char)(s[i] + '0') + res;
        if(res == "")    res = "0";
        return res;
    }

    bign operator + (const bign b) const
    {
        bign c;
        c.len = 0;
        for(int i = 0, g = 0; g || i < max(len, b.len); ++i)
        {
            int x = g;
            if(i < len)    x += s[i];
            if(i < b.len)    x += b.s[i];
            c.s[c.len++] = x % 10;
            g = x / 10;
        }
        return c;
    }

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

istream& operator >> (istream& in, bign& x)
{
    string s;
    in >> s;
    x = s.c_str();
    return in;
}

ostream& operator << (ostream &out, const bign& x)
{
    out << x.str();
    return out;
}

const int maxn = 2000;
bign d[maxn + 10];
int c[] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};

int main(void)
{
    #ifdef LOCAL
        freopen("11375in.txt", "r", stdin);
    #endif

    memset(d, 0, sizeof(d));
    d[0] = 1;
    for(int i = 0; i <= maxn; ++i)
        for(int j = 0; j < 10; ++j)
            if(!(i==0 && j==0) && i+c[j]<=maxn)
                d[i+c[j]] += d[i];
    for(int i = 2; i <= maxn; ++i)
        d[i] += d[i-1];
    int x;
    while(scanf("%d", &x) == 1)
    {
        if(x < 6)    cout << d[x] << endl;
        else    cout << d[x] + 1 << endl;
    }
    return 0;
}