URAL1002——DP——K-based Numbers. Version 2（未AC）

Description

Let’s consider  K-based numbers, containing exactly  N digits. We define a number to be valid if its  K-based notation doesn’t contain two successive zeros. For example:

• 1010230 is a valid 7-digit number;
• 1000198 is not a valid number;
• 0001235 is not a 7-digit number, it is a 4-digit number.

Given two numbers  N and  K, you are to calculate an amount of valid  K based numbers, containing  N digits.

You may assume that 2 ≤  K ≤ 10;  N ≥ 2;  N +  K ≤ 1800.

Input

The numbers  N and  K in decimal notation separated by the line break.

Output

The result in decimal notation.

Sample Input

input	output
2 10	90

 大意：增大了数据，范围，就是求1到n位数用k进制形式的排除两个零在一起的一共多少情况

这题目有毒orz。。。。。超内存怎么搞。。。不过学习了bign（重载），妈妈再也不用担心我不会高精度了~~~

MLE代码。。

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<string>

using namespace std;

const int maxn = 2000;

struct bign

{

    int len ,s[maxn];

    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[maxn];

        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;

    }

};

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;

}

bign dp[1900][11];

int n,k;

int main()

{

    scanf("%d%d",&n,&k);

    for(int i = 1; i < k ;i++)

        dp[1][i] = 1;

        for(int i = 2; i <= n ;i++){

            for(int j = 0 ; j < k ; j++){

                for(int l = 0 ; l < k;l++){

                    if(!j && !l) continue;

                    dp[i][l] = dp[i][l] + dp[i-1][j];

                }

            }

        }

        bign ans = 0;

        for(int i = 0 ; i < k ;i++)

            ans = ans + dp[n][i];

        cout<<ans<<endl;

     return 0;

}