Ideas: Digital $ DP $
Submission: 5 (in fact, before A before, but adjusted the original program this is twice the AC.)
answer:
We are determined $ sum (x) = i $, for a $ i $, there are several $ x $, then we can quickly solve power.
As required by the number of the DP digit $ $ like.
#include<cstdio> #include<iostream> #include<cstring> #define ull unsigned long long #define ll long long #define R register ll using namespace std; #define pause (for(R i=1;i<=10000000000;++i)) #define In freopen("NOIPAK++.in","r",stdin) #define Out freopen("out.out","w",stdout) namespace Fread { static char B[1<<15],*S=B,*D=B; #ifndef JACK #define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++) #endif inline ll g() { R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-1:fix; if(ch==EOF) return EOF; do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix; } inline bool isempty(const char& ch) {return (ch<=36||ch>=127);} inline void gs(char* s) { register char ch; while(isempty(ch=getchar())); do *s++=ch; while(!isempty(ch=getchar())); } } using Fread::g; using Fread::gs; namespace Luitaryi { const int N=51,M=1e7+7; ll n; int len,num[N]; ll f[N][N]; inline int qpow(ll a,ll p) { R ret=1; a%=M; for(; P; P >> = . 1 , (A * = A)% = M) IF (P & . 1 ) (RET * = A)% = M; return RET; } inline LL DFS ( int L, BOOL UL, int C, int D) { // L: length, ul: upper bound tag, c: count the number 1, d: the number of a request (i.e., at this time we make SUM (X) = D) IF (L! ) return C == D; IF (UL && ~ F [L] [C])! return F [L] [C]; R & lt Lim = (UL NUM [L]:? . 1 ), CNT = 0 ; for (R & lt = I 0 ; I <= Lim; ++ I) CNT + = DFS (L- . 1,ul&&i==lim,c+i,d); return ul?cnt:f[l][c]=cnt; } inline int solve(ll n) { R ans=1; for(;n;n>>=1) num[++len]=n&1; for(R i=1;i<=len;++i) memset(f,0xff,sizeof(f)), ans=(ans*qpow(i,dfs(len,true,0,i)))%M; return ans; } inline void main() { n=g(); printf("%d\n",solve(n)); } } signed main() { Luitaryi::main(); return 0; }
2019.07.21