2016沈阳网络赛 odd-even number

时间：2024-01-11 20:16:38

odd-even number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description
For a number,if the length of continuous odd digits is
even and the length of continuous even digits is odd,we call it odd-even
number.Now we want to know the amount of odd-even number between
L,R(1<=L<=R<= 9*10^18).
Input
First line a t,then t cases.every line contains two
integers L and R.
Output
Print the output for each case on one line in the
format as shown below.
Sample Input
2
1 100
110 220
Sample Output
Case #1: 29 Case #2: 36
分析：数位dp；
　　　dp[i][j][k]三维分别代表位置，当前数奇偶性，个数奇偶性；
　　　注意前导0的特判；
代码：
#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <climits>

#include <cstring>

#include <string>

#include <set>

#include <bitset>

#include <map>

#include <queue>

#include <stack>

#include <vector>

#define rep(i,m,n) for(i=m;i<=n;i++)

#define mod 1000000007

#define inf 0x3f3f3f3f

#define vi vector<int>

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define ll long long

#define pi acos(-1.0)

#define pii pair<int,int>

#define sys system("pause")

const int maxn=1e5+;

const int N=5e4+;

const int M=N**;

using namespace std;

inline ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}

inline ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}

inline void umax(ll &p,ll q){if(p<q)p=q;}

inline void umin(ll &p,ll q){if(p>q)p=q;}

inline ll read()

{

    ll x=;int f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

int n,m,k,t,num[],pos,cas;

ll dp[][][],p,q;

ll dfs(int pos,int x,int y,int z,int k)

{

    if(pos<)return k&&(x^y)==;

    if(z&&k&&dp[pos][x][y]!=-)return dp[pos][x][y];

    int now=z?:num[pos],i;

    ll ret=;

    rep(i,,now)

    {

        if(k&&(x^(i&))==&&(x^y)==)continue;

        if(!i&&!k)ret+=dfs(pos-,x,y,z||i<num[pos],k||i);

        else if((i&)==x)ret+=dfs(pos-,x,y^,z||i<num[pos],k||i);

        else ret+=dfs(pos-,x^,,z||i<num[pos],k||i);

    }

    return z&&k?dp[pos][x][y]=ret:ret;

}

ll gao(ll x)

{

    pos=;

    while(x)num[pos++]=x%,x/=;

    return dfs(pos-,,,,);

}

int main()

{

    int i,j;

    memset(dp,-,sizeof(dp));

    scanf("%d",&t);

    while(t--)

    {

        scanf("%lld%lld",&p,&q);

        printf("Case #%d: %lld\n",++cas,gao(q)-gao(p-));

    }

    return ;

}

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