NOJ1545---New Year 2014(数位dp)

问题描述

In the New Year 2014, Xiao Ming is thinking about the question: give two integers N and K, Calculate the number of the numbers of satisfy the following conditions:

1. It is a positive integer and is not greater than N.

2. Xor value of its all digital is K.

For example N = 12, K = 3, the number of satisfy condition is 3 and 12 (3 = 3, 1 ^ 2 = 3).

In order to let Xiao Ming happy in the New Year 2014, can you help him?
输入
There are multiple test cases, each test sample contains two positive integers N and K (0 <= N, K < 10 ^ 18), End to file.
输出
For each case, output the number of the numbers of satisfy condition in one line.
样例输入

12 3
999 5
12354 8

样例输出

2
76
662

提示

无

来源

宁静致远 @HBMY

操作

曾几何时,这道题根本不会,最近oj修好了,偶然看到这道题,就来弥补一下缺憾
很水的数位dp 显然个位数异或顶多到15

/************************************************************************* > File Name: NOJ1545.cpp > Author: ALex > Mail: [email protected] > Created Time: 2015年04月14日 星期二 15时21分34秒 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

LL dp[30][30];
int bit[30];

LL dfs(int cur, int _xor, bool flag, bool zero)
{
    if (cur == -1)
    {
        if (zero)
        {
            return 0;
        }
        return _xor == 0;
    }
    if (!flag && ~dp[cur][_xor])
    {
        return dp[cur][_xor];
    }
    LL ans = 0;
    int end = flag ? bit[cur] : 9;
    for (int i = 0; i <= end; ++i)
    {
        if (zero && !i)
        {
            ans += dfs(cur - 1, _xor, flag && (i == end), 1);
        }
        else
        {
            ans += dfs(cur - 1, _xor ^ i, flag && (i == end), 0);
        }
    }
    if (!flag)
    {
        dp[cur][_xor] = ans;
    }
    return ans;
}

LL calc(LL n, LL k)
{
    int cnt = 0;
    while (n)
    {
        bit[cnt++] = n % 10;
        n /= 10;
    }
    return dfs(cnt - 1, k, 1, 1);
}

int main()
{
    memset(dp, -1, sizeof(dp));
    LL n, k;
    while (cin >> n >> k)
    {
        if (k > 15)
        {
            cout << 0 << endl;
            continue;
        }
        int low = 0;
        if (k == 0)
        {
            ++low;
        }
        cout << calc(n, k) - low << endl;
    }
    return 0;
}