[1545] New Year 2014（数位DP，现放标程，待看）

1、http://acm.nbut.edu.cn/Problem/view.xhtml?id=1545

2、题目

[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 Nand 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 3999 512354 8

样例输出

276662

提示

无

来源

宁静致远 @HBMY

 

3、比赛标程

简单的数位DP, dp[pos][statu] (statu 为异或值)!
本题有个trick, k等于0的时候要将答案减一(题目中要求数都为正的).

#include<cstdio>#include<cstring>#include<iostream>using namespace std;#define lint long longint bit[30];lint dp[30][20];lint dfs(int pos,int x,bool limit){    if(pos<0)    {        if(limit)return 0;        else return x==0;    }    if(!limit&&dp[pos][x]!=-1)return dp[pos][x];    int end=(limit==0)?9:bit[pos];    lint ans=0;    for(int i=0;i<=end;++i)    {        ans+=dfs(pos-1,i^x,limit&&(i==end));    }    dp[pos][x]=ans;    return ans;}lint solve(lint n,int x){    int len=-1;    memset(dp,-1,sizeof(dp));    while(n)    {        bit[++len]=n%10;        n/=10;    }    return dfs(len,x,1);}int main(){    //freopen("G:\\1.in","r",stdin);    //freopen("G:\\1.out","w",stdout);    lint n,x;    while(cin>>n>>x)    {        if(x>15)        {            puts("0");            continue;        }        lint ans=solve(n+1,x);        if(x==0)--ans;        printf("%lld\n",ans);    }    return 0;}