UVALive

题目链接:https://cn.vjudge.net/problem/UVALive-7344
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You have N cards and each has an unique number between 1 and N written on it. In how many
ways can you select a non-empty subset of the cards such that the number written on any two of your selected cards don’t have any common digits?
For example, when N = 12, {1, 2, 3}, {2, 11}, {3, 4, 5, 6, 7, 8, 9, 12} are some valid selections. But
{1, 2, 10}, {2, 5, 12} are not allowed.

Input
The first line of the input contains an integer T (T≤15) which is the number of test cases. Each of
the following T lines denote a test case, containing an integer N (1≤N<109).

Output
For each test case, output the case number followed by the number of subsets modulo 1000000007.

Sample Input
2
3
12

Sample Output
Case 1: 7
Case 2: 1151
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题目大意：
让你在[1,N]内选择一些数构成一个集合,使得集合中任意两个数没有相同的数字,(但一个数可以由相同的数字比如11.22),问你合法的集合的个数

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首先能确定的是集合中数最多不超过10个，那么可以枚举来

先选择一个，在选择第二个，保证其两者不没有相同位即可

显然这个过程是可以进行dp的
用长度为10的二进制表示0~9.然后进行转移即可

设dp[i][j],表示取到第i个数是选择了j(二进制)对应的这些数字

从而有dp[i][j]=∑dp[i−1][k]∗A(表示k\xorj这个状态下的数的个数)

显然有一个O(10*1024*1024)即可

但是对于A(表示k\xorj这个状态下的数的个数)我们应该怎么求呢

显然是预处理啊 ，
通过数位dp进行统计，预处理出每种状态下的数的个数。

附本题代码
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LL dp_nex[15][(1<<12)];int n;int num[12],len;int dp[12][1111];int f[1111];int dfs(int pos,int limit,int na,int &nb){    if(pos<0) return na==nb;    int &t=dp[pos][na];    if(!limit&&t!=-1) return t;    int endi = 9;if(limit) endi=num[pos];    int res = 0;    for(int i=0;i<=endi;i++){        if( (na|i) == 0 )            res+=dfs(pos-1,endi==i&&limit,na,nb);        else if(nb&(1<<i))            res+=dfs(pos-1,endi==i&&limit,na|(1<<i),nb);    }    if(!limit) t=res;    return res;}void solve(int x){    for(len=0;x;x/=10) num[len++]=x%10;    f[0]=f[1]=0;    for(int i=2;i<(1<<10);i++){        memset(dp,-1,sizeof(dp));        f[i]=dfs(len-1,1,0,i);    }}LL qmod(LL a,LL b){    LL res = 1;    while(b){        if(b&1) res=res*a%MOD;        b>>=1;  a  =a  *a%MOD;    }    return res;}int main(){    int t,kcase=0;scanf("%d",&t);    while(t--){        int n;        scanf("%d",&n);        memset(dp_nex,0,sizeof(dp_nex));        solve(n);        dp_nex[0][0]=1;        int endi=(1<<10);        for(int i=1;i<=10;i++){            for(int j=0;j<endi;j++){                for(int k=0;k<endi;k++){                    if(j==k)continue;                    if(((j&k)==k))                        dp_nex[i][j]=(dp_nex[i][j]+dp_nex[i-1][k]*f[j-k])%MOD;                }            }        }        LL output=0;        for(int i=1;i<=10;i++){            int tmp=1;            for(int j=1;j<=i;j++) tmp*=j;            for(int j=0;j<endi;j++)                output=(output+qmod(tmp,MOD-2)*dp_nex[i][j])%MOD;        }        printf("Case %d: %lld\n",++kcase,output);    }    return 0;}