SRM 597 Div 2 1000 LittleElephantAndSubset

Problem Statement
	Little Elephant from the Zoo of Lviv has a set S. The set S contains the integers 1, 2, 3, ...,N. He considers some non-empty subset of S cool if we can write down all elements of the subset by using each of the digits 0-9 at most once. Examples: The subsets {12,345,67890} and {47,109} are cool. The subset {147,342} is not cool because the digit 4 is used twice. The subset {404} is not cool for the same reason. You are given the int N. Let X be the total number of non-empty cool subsets of S. Return the value (X modulo 1,000,000,007).
Definition
	Class: LittleElephantAndSubset Method: getNumber Parameters: int Returns: int Method signature: int getNumber(int N) (be sure your method is public)
Constraints
-	N will be between 1 and 1,000,000,000 (10^9), inclusive.
Examples
0)
	3 Returns: 7 All 7 non-empty subsets are cool in this case: {1, 2, 3} {1, 2} {1, 3} {2, 3} {1} {2} {3}
1)
	10 Returns: 767 In this case, the total number of non-empty subsets is 2^10-1 = 1023. The only not cool subsets are those that contain both 1 and 10. There are 2^8 = 256 of them. Thus, the answer is 1023-256 = 767.
2)
	47 Returns: 25775
3)
	4777447 Returns: 66437071 数位DP，先预处理出cnt[S]表示选0..9状态为S时能够得到的不超过N的数字个数（不能有前导0）。 注意数字本身也不能有重复数字。 然后就是背包。 F[S]=状态为S时{1,2,3,...N}的子集数。 状态转移方程：F[i|S]+=F[i]*cnt[S];    i&S==0 ANS=SIGMA(F[S])    S>0 #include <vector> #include <list> #include <map> #include <set> #include <deque> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <ctime> #include <cstring> using namespace std; #define ll long long const int mod=1e9+7; int cnt[1884]; bool vis[33]; int bit[334]; int dp[22][3322]; int now; int dfs(int len,int s,bool lim,bool pos) { if(!len)return pos&&s==now; if(!lim&&pos&&dp[len][s]!=-1)return dp[len][s]; int ed=lim?bit[len-1]:9; int ans=0; for(int i=0;i<=ed;i++) { if(vis[i]) { if(s&(1<<i))continue; int news; if(!pos&&!i)news=s; else news=s|(1<<i); ans+=dfs(len-1,news,lim&&i==ed,pos||i); } } if(!lim&&pos)dp[len][s]=ans; return ans; } ll f[2334]; class LittleElephantAndSubset { public: int getNumber(int); }test; int LittleElephantAndSubset::getNumber(int N) { int tot=0; for(int t=N;t;t/=10)bit[tot++]=t%10; memset(cnt,0,sizeof cnt); for(int i=0;i<1024;i++) { memset(vis,0,sizeof vis); vis[0]=1; for(int j=0;j<10;j++)if(i>>j&1)vis[j]=1; now=i; memset(dp,-1,sizeof dp); cnt[i]=dfs(tot,0,1,0); ///int dfs(int len,int s,bool lim,bool pos) } memset(f,0,sizeof f); f[0]=1; for(int s=1;s<1024;s++)if(cnt[s]) { for(int i=0;i<1024;i++)if(!(i&s)) { f[i|s]+=(f[i]*cnt[s]%mod); f[i|s]%=mod; } } int ans=0; for(int i=1;i<1024;i++) { ans+=f[i]; if(ans>=mod)ans-=mod; } return ans; } int main() { int n; while(cin>>n) { int ans=test.getNumber(n); cout<<ans<<endl; } } ///<%:testing-code%> //Powered by [KawigiEdit] 2.0!