cf 611 B ||codeforces goodbye 2015 B. New Year and Old Property (数学或者数位dp)

2016-01-01 · 2 min read · dp math number theory 位运算数位dp

http://codeforces.com/contest/611/problem/B 题意：问a到b（1E18），二进制表示中只有一个0的数有多少个。思路：~~这么大的数。。。不是有循环节就是math problems.~~ UD:20160318讲道理还有可能是数位dp好不好。。。我们发现可以很容易得算出1到x的二进制表示中只有一个0 的数有多少个。

problem solved.

20160318update:学了数位dp后又看到这题。。。这题显然是数位dp啊。。。亏我找规律搞了出来2333.

后面附上数位dp方法AC的代码

/* ***********************************************
Author :111qqz
Created Time :2015年12月30日 星期三 22时49分02秒
File Name :code/cf/goodbye2015/B.cpp
************************************************ */

 1#include <cstdio>
 2#include <cstring>
 3#include <iostream>
 4#include <algorithm>
 5#include <vector>
 6#include <queue>
 7#include <set>
 8#include <map>
 9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair

 1using namespace std;
 2const int N=1E4+7;
 3LL a,b;
 4LL p[N];
 5LL c[N];
 6LL cal( LL x)
 7{
 8    return ((x-1LL)*x)/2LL;
 9}
10LL solve (LL x)
11{
12    if (x==0LL) return 0;
13    LL res= 0LL;
14    LL cnt = 0LL;
15    LL xx = x;
16    while (xx)
17    {
18	cnt++;
19	p[cnt] = xx%2LL;
20	xx/=2LL;
21    }
22    ms(c,0);
23    res+=cal(cnt-1LL);
24    LL tmp = (1LL<<cnt)-1LL;
25    for ( LL i = 0 ; i <cnt-1 ; i++)
26    {
27	LL happ = 1LL<<i;
28	c[i]=tmp-happ;
29    }

    sort(c,c+cnt-1);

1    for ( LL i = 0  ; i< cnt -1 ; i++)
2    {
3	if (x>=c[i]) res++;
4    }

1    return res;
2}
3int main()
4{

1	cin>>a>>b;
2	LL ans = solve(b)-solve(a-1LL);
3	cout<<ans<<endl;

1  #ifndef ONLINE_JUDGE  
2  fclose(stdin);
3  #endif
4    return 0;
5}

数位dp的方法：

/* ***********************************************
Author :111qqz
Created Time :2016年03月18日 星期五 16时33分04秒
File Name :code/cf/problem/611B.cpp
************************************************ */

 1#include <cstdio>
 2#include <cstring>
 3#include <iostream>
 4#include <algorithm>
 5#include <vector>
 6#include <queue>
 7#include <set>
 8#include <map>
 9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair

1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6LL l,r;
7int digit[80];
8LL dp[80][80];

1LL dfs( int pos,int cnt,bool limit,bool prehasnonzero)
2{
3    if (pos==0) return cnt==1;
4    if (!limit&&prehasnonzero&&dp[pos][cnt]!=-1) return dp[pos][cnt];

 1    int mx = limit?digit[pos]:1;
 2    LL res = 0LL ;
 3    if (prehasnonzero)
 4    {
 5	for ( int i =  0 ; i <= mx ; i++)
 6	{
 7	    res+=dfs(pos-1,i==0?cnt+1:cnt,limit&&i==mx,true);
 8	}
 9    }
10    else
11    {
12	for ( int i = 0 ; i <= mx; i++)
13	{
14	    res+=dfs(pos-1,0,limit&&i==mx,i==0?false:true);
15	}
16    }

1    if (!limit&&prehasnonzero) dp[pos][cnt] = res;
2    return res;
3}
4LL solve  ( LL n)
5{
6    int len =  0 ;
7    ms(digit,0);

 1    while (n)
 2    {
 3	digit[++len] = n % 2;
 4	n /= 2;
 5    }
 6    return dfs(len,0,true,false);
 7}
 8int main()
 9{
10	#ifndef  ONLINE_JUDGE 
11	freopen("code/in.txt","r",stdin);
12  #endif
13	ms(dp,-1);
14	cin>>l>>r;
15	//cout<<"solve:"<<solve()<<endl;
16	LL ans = solve(r) - solve (l-1);
17	cout<<ans<<endl;

1  #ifndef ONLINE_JUDGE  
2  fclose(stdin);
3  #endif
4    return 0;
5}