简单递推系列 uva习题

uva 11375 高精度加递推

主要是高精度加递推（主要思想还是参考liurujia书上）自己想了半天没有想出来。就不浪费时间了，理解后果断直接上模版。还犯了一个小错误wa一次0.0

//
//  BigNum.cpp
//  By Zhang xiaohao
//
//
#include <iostream>
#include <cstdio>
#include <cstring>
#define maxn 1010
using namespace std;

struct BigNum{
    char str[maxn];     //数字反着存方便运算
    int len;

    BigNum():len(0){
        memset(str, 0, sizeof str);
    }

    BigNum operator+(BigNum &num)
    {
        BigNum ret;
        for(int i=0, g=0; g || i<max(len, num.len); i++){
            int ts = g;
            if(i<len) ts+=(int)str[i]-'0';
            if(i<num.len) ts+=(int)num.str[i]-'0';
            g = ts/10;
            ret.str[ret.len++] = ts%10+'0';
        }
        ret.str[ret.len] = '';
        return ret;
    }

    BigNum operator+(int n)
    {
        BigNum num, ret;
        sprintf(num.str, "%d", n);
        num.len = strlen(num.str);
        for(int i=0; i<num.len/2; i++){
            swap(num.str[i], num.str[num.len-i-1]);
        }
        ret = num+*this;
        return ret;
    }

    BigNum operator=(char c[]){
        BigNum ret;
        ret.len = strlen(c);
        for(int i=ret.len-1; i>=0; i--){
            ret.str[i] = c[i];
        }
        ret.str[len] = '';
        return ret;
    }

    BigNum operator=(int num){
        sprintf(this->str, "%d", num);
        this->len = strlen(this->str);
        for(int i=0; i<this->len/2; i++){
            swap(this->str[i], this->str[this->len-i-1]);
        }
        return *this;
    }
};

ostream& operator <<(ostream &out, const BigNum& x){
    for(int i=x.len-1; i>=0; i--){
        out << x.str[i];
    }
    return out;
}

istream& operator >>(istream &in, BigNum& x){
    in >> x.str;;
    x.len = strlen(x.str);
    for(int i=0; i<x.len/2; i++){
        swap(x.str[i], x.str[x.len-i-1]);
    }
    return in;
}

int n, c[] = {6,2,5,5,4,5,6,3,7,6};
BigNum dp[5010], sum[5010];

void init()
{
    for(int i=0; i<maxn; i++)dp[i] = 0;
    for(int i=0; i<maxn; i++)sum[i] = 0;
    dp[0] = 1;
    sum[0] = 0;
    for(int i = 0; i<=2000; i++){
        for(int j=0; j<10; j++){
            if(!(i==0 && j==0) && i+c[j]<=2000) dp[i+c[j]] = dp[i]+dp[i+c[j]];
        }
    }
    for(int i=1; i<=2000; i++){
        sum[i] = sum[i-1]+dp[i];
        if(i==6) sum[i] = sum[i]+1;
    }
}

void debug()
{
    for(int i=0; i<10; i++)
    cout << dp[i] << endl;
}

int main()
{
//    freopen("in.txt", "r", stdin);

    init();
    while(scanf("%d", &n)!=EOF)
    {
        cout << sum[n] << endl;
    }
    return 0;
}