Fighting For 2014 (One University One Question) Round II [E] New Year 2014

• [E] New Year 2014

• 时间限制: 2000 ms　内存限制: 262144 K
• 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

很久没做题了，对于这种最简单的数位dp都做不了！

题意：求1~n的数中符合每一数位的异或为k的个数。

题解：最简单的数位dp，不过要懂得a = b1^b2^……bm   推出  a^b1^b2……bm = 0；

所以我们dfs时dp[i][j]代表着有i位和i位一下的所有整数的数位异或值为j的个数。

然后就一步步递归下去。。

不过这些都是包括零，所以还要减去0.。

#include<cstdio>
#include<cstring>
#include<math.h>
#include<algorithm>
#include<vector>
#include<queue>
#include<iostream>
#include<cstdlib>

using namespace std;

int Max(int a,int b)
{
	return a>b?a:b;
}
long long dp[25][20];
long long n,k;
int num[20];
int tran(long long n)
{
	int i = 0;
	while(n)
	{
		num[i++] = n%10;
		n/=10;
	}
	return i;
}
long long DFSdp(int b,int s,int r,int doing)
{
	
	if(b==0 ) 
	{
		if(r==0)
			return 1;
		else
			return 0;
	}
	if(!doing && dp[b][r]!=-1) 
	{
		return dp[b][r];
	}
	int end = doing==1?num[b-1]:9;
	long long ans = 0;
	for(int i=0;i<=end;i++)
	{
		ans += DFSdp(b-1,i,r^i,doing&&i==end);
	}
	if(!doing)
		dp[b][r] = ans;
	return ans;
}
int main()
{
	memset(dp,-1,sizeof(dp));
	
	while(~scanf("%lld%lld",&n,&k))
	{
		int b = tran(n);
		if(k>15) 
		{
			printf("0\n");
			continue ;
		}
		long long ans = DFSdp(b,0,(int)k,1);
		if(k==0)
			ans--;
		printf("%lld\n",ans);
	}
	
    return 0;
}

Fighting For 2014 (One University One Question) Round II [E] New Year 2014

原文：http://blog.csdn.net/min_lala/article/details/19208753