限于本人DP处于背包--水平，不组声（就是对神奇的数位DP，找出j个1的有多少个，然后j的多少个次方就行）
1 #include<bits/stdc++.h> 2 #define N 100005 3 #define LL long long 4 #define inf 0x3f3f3f3f 5 using namespace std; 6 inline LL ra() 7 { 8 LL x=0,f=1; char ch=getchar(); 9 while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();} 10 while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();} 11 return x*f; 12 } 13 const LL mod=10000007LL; 14 int l,wei[100]; 15 LL c[100][100],n,ans; 16 void pre() 17 { 18 for (int i=0; i<=60; i++) c[i][0]=1; 19 for (int i=1; i<=60; i++) 20 for (int j=1; j<=i; j++) 21 c[i][j]=c[i-1][j-1]+c[i-1][j]; 22 } 23 LL ksm(LL a, LL b) 24 { 25 LL sum=1; a%=mod; 26 for(;b;b>>=1,a=a*a%mod) 27 if (b&1LL) sum=sum*a%mod; 28 return sum; 29 } 30 LL solve(int x) 31 { 32 LL sum=0; 33 for (int i=l; i>=1; i--) 34 { 35 if (wei[i]==1) 36 sum+=c[i-1][x--];//计算的时候默认i位是0，计算完后默认i位是1，，应该是这样子的，，不会dp不会dp，，， 37 //这样求出来的应该就是最小值了233 38 if (x<0) break; 39 } 40 return sum; 41 } 42 int main() 43 { 44 pre(); n=ra()+1; 45 while (n) 46 { 47 wei[++l]=n&1; 48 n>>=1; 49 } 50 ans=1LL; 51 for (int i=1; i<=l; i++) 52 ans=ans*ksm(i,solve(i))%mod; 53 printf("%lld ",ans); 54 return 0; 55 }