bzoj 3209 数位DP+欧拉定理

枚举1的个数,统计有那么多1的数的个数

 /**************************************************************
     Problem: 3209
     User: idy002
     Language: C++
     Result: Accepted
     Time:0 ms
     Memory:844 kb
 ****************************************************************/

 #include <cstdio>
 #define Mod 10000007
 #define Phi 9988440

 typedef long long dnt;

 dnt n;
 dnt dp[][][];
 int tb;

 void dodp() {
     for( tb=; tb>=; tb-- )
         if( (n>>tb)& ) break;
     dp[tb][][] = ;
     dp[tb][][] = ;
     for( int b=tb-; b>=; b-- ) {
         for( int c=; c<=tb-b+; c++ ) {
             dp[b][][c] = dp[b+][][c-((n>>b)&)];
             dp[b][][c] = (dp[b+][][c] + dp[b+][][c-]) % Phi;
             if( (n>>b)& ) dp[b][][c] = (dp[b][][c] + dp[b+][][c]) % Phi;
         }
     }
 }
 dnt mpow( dnt a, dnt b ) {
     dnt rt;
     for( rt=; b; b>>=,a=(a*a)%Mod )
         if( b& ) rt=(rt*a)%Mod;
     return rt;
 }
 int main() {
     scanf( "%lld", &n );
     dodp();
     dnt ans = ;
     for( int c=; c<=tb+; c++ )
         ans = (ans * mpow(c,dp[][][c]+dp[][][c])) % Mod;
     printf( "%lld\n", ans );
 }