[BZOJ 3209] 花神的数论题

[题目链接]

        https://www.lydsy.com/JudgeOnline/problem.php?id=3209

[算法]

         数位DP

         记f[i][j][k]表示i位， 最高为为j ， 有k个1的二进制数有多少个

         然后 ， 计算1-N中 ， 出现i个1的数有多少个

         用快速幂将答案乘起来 ， 即可

[算法]

        

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL MAXN = 1e15 + 10;
const int MAXLOG = 50;
const int P = 10000007;

LL n;
LL ans[MAXLOG + 10];
LL f[MAXLOG + 10][2][MAXLOG + 10];

template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); }
template <typename T> inline void chkmax(T &x , T y) { x = max(x , y); }
template <typename T> inline void read(T &x)
{
   T f = 1; x = 0;
   char c = getchar();
   for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
   for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
   x *= f;  
}
inline void preprocess()
{
    f[1][0][0] = f[1][1][1] = 1;
    for (int i = 1; i < MAXLOG; i++)
    {
        for (int j = 0; j < 2; j++)
        {
            for (int k = 0; k <= i; k++)
            {
                f[i + 1][1][k + 1] += f[i][j][k];
                f[i + 1][0][k] += f[i][j][k];
            }
        }
    }
}
inline void calc(LL n)
{
    int len = 0;
    static int a[MAXLOG];
    LL ret = 0;
    while (n > 0)
    {
        a[++len] = n & 1;
        n >>= 1;
    }    
    reverse(a + 1 , a + len + 1);
    int cnt = 0;
    for (int i = 1; i <= len; i++)
    {
        if (a[i] == 1) 
        {
            for (int j = 0; j <= len - i + 1; j++)
                ans[cnt + j] += f[len - i + 1][0][j];
            ++cnt;        
        } else continue;
    }
}
inline int exp_mod(LL a , LL n)
{
    LL res = 1 , b = a;
    while (n > 0)
    {
        if (n & 1) res = 1LL * res * b % P;
        b = 1LL * b * b % P;
        n >>= 1;
    }    
    return res;
}

int main()
{
    
    preprocess();
    read(n);
    calc(n + 1);
    int res = 1;
    for (int i = 1; i < MAXLOG; i++) res = 1LL * res * exp_mod(i , ans[i]) % P;
    printf("%d
" , res);
    
    return 0;
}