（bzoj 3209 花神的数论题）<>

传送门

Solution

将数字看成二进制
f[i][j]表示前i为中有j个1的数的总数
f[i][j]=f[i−1][j−1]+f[i−1][j]
然后逐位统计即可

Code

// by spli#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#define LL long longusing namespace std;const LL p=10000007;LL n;LL f[70][70];int bit[70],num=0;LL exp(LL a,LL b){    LL ret=1,k=a%p;    while(b){        if(b&1) (ret*=k)%=p;        b>>=1;        (k*=k)%=p;    }    return ret%p;}LL get(int x){    LL s=0;    for(int i=num;i>=1;--i){        if(bit[i]==1){            s+=f[i-1][x];            x--;        }        if(x<0) break;    }    return s;}int main(){    scanf("%lld",&n);    n++;    for(int i=0;i<=60;++i) f[i][0]=1;    for(int i=1;i<=60;++i)        for(int j=1;j<=i;++j)            f[i][j]=f[i-1][j]+f[i-1][j-1];//取模会WA     while(n){        bit[++num]=n&1;        n>>=1;    }    LL ans=1;    for(int i=1;i<=num;++i)        (ans*=exp((LL)i,get(i)))%=p;    printf("%lld\n",ans);    return 0;}