洛谷 P4317 花神的数论题(组合数)

题面

luogu

题解

组合数

枚举有多少个\(1\),求出有多少种数

扫描\(n\)的每一位\(1\), 强制选\(0\)然后组合数算一下有多少种方案

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;
template<class T> inline void read(T &x) {
	x = 0; RG char c = getchar(); bool f = 0;
	while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
	while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
	x = f ? -x : x;
	return ;
}
template<class T> inline void write(T x) {
	if (!x) {putchar(48);return ;}
	if (x < 0) x = -x, putchar('-');
	int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
	for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
}
const int N = 55, Mod = 10000007;
LL C[N][N], cnt, a[N];
inline LL Pow(int a, LL b) {
	LL s = 1;
	for (LL x = a; b; b >>= 1, x = x*x%Mod) if (b&1) s = s*x%Mod;
	return s;
}
int main() {
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
	for (int i = 0; i <= 50; i++) C[i][i] = C[i][0] = 1;
	for (int i = 2; i <= 50; i++)
		for (int j = 1; j < i; j++)
			C[i][j] = C[i-1][j-1]+C[i-1][j];
	LL n; read(n);
	for (int i = 50; i >= 0; i--) {
		if ((n>>i)&1) {
			for (int j = 1; j <= i; j++)
				a[j+cnt] += C[i][j];//至少选1个的方案
			a[++cnt]++;//不选的方案(必须分开算,不然会有重复计算)
		}
	}
	LL ans = 1;
	for (int i = 1; i <= 50; i++)
		ans = ans*Pow(i, a[i])%Mod;
	printf("%lld\n", ans);
	return 0;
}