[1545] New Year 2014

• [1545] 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

    DP的思路倒是很容易的就想出来了，但是没想到一个小细节，让我查了一天的数据，。。。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
__int64 dp1[16][20],dp2[16][20];
char s1[20],s2[20];
int main()
{
    //freopen("data.in","r",stdin);
    memset(dp1,0,sizeof(dp1));
    memset(dp2,0,sizeof(dp2));
    for(int i=0;i<=9;i++)
    {
        dp1[i][1] = dp2[i][1] = 1;
    }
    dp1[0][0] = dp2[0][0] = 1;
    for(int i=2;i<=19;i++)
    {
        for(int j=1;j<=9;j++)
        {
            for(int k = 0;k<=15;k++)
            {
                int u = k^j;
                dp1[u][i]+=dp2[k][i-1];  //不带前导0
            }
        }
        for(int j=0;j<=9;j++)
        {
            for(int k=0;k<=15;k++)
            {
                int u = k^j;
                dp2[u][i]+=dp2[k][i-1];   //带前导0
            }
        }
    }
    while(scanf("%s %s",s1,s2)!=EOF)
    {
        int l2 = strlen(s2);
        int l1 = strlen(s1);
        if(l2>2)
        {
            printf("0\n");
            continue;
        }
        int s= 0;
        for(int i=0;i<=l2-1;i++)
        {
            s = s*10+s2[i]-‘0‘;
        }
        if(s>15)
        {
            printf("0\n");
            continue;
        }
        __int64 res = 0;
        for(int i=1;i<=l1-1;i++)
        {
            res+=dp1[s][i];
        }
        int tag = 0;
        for(int i=0;i<=l1-1;i++)
        {
            int dig = s1[i]-‘0‘;
            int t;
            if(i==0)
            {
                t =1;
            }else
            {
                t = 0;
            }
            for(int j=t;j<=dig-1;j++)
            {
                for(int k=0;k<=15;k++)
                {
                    int x = tag^j;
                    x = x^k;
                    if(x==s)
                    {
                        res+=dp2[k][l1-1-i];
                    }
                }
            }
            tag = tag^dig;
        }
        if(l1==1&&s==0)
        {
            res++;
        }else
        {
            tag = 0;
            for(int i=0;i<=l1-1;i++)
            {
                int dig = s1[i]-‘0‘;
                tag = tag^dig;
            }
            if(tag==s)
            {
                res++;
            }
        }
        if(s==0)
        {
            res-=1;
        }
        printf("%I64d\n",res);
    }
    return 0;
}

[1545] New Year 2014

原文：http://blog.csdn.net/yongxingao/article/details/19188809