洛谷 P2518 [HAOI2010]计数 (组合数)

题面

luogu

题解

本来想练数位dp的,结果又忍不住写了组合数..

去掉一个(0)可以看作把(0)移到前面去

那么题目转化为 (n)有多少个排列小于(n)

强制某一位比(n)的对应位置上的数小, 后面方案组合数算一下即可

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 ;
}
char s[55];
int a[55], b[10], C[55][55];

int main() {
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
	scanf("%s", s);
	int n = strlen(s);
	for (int i = 0; i < n; i++)
		a[i+1] = s[i]-'0', b[a[i+1]]++;
	LL ans = 0;
	for (int i = 0; i <= n; i++) C[i][i] = 1, C[i][0] = 1;
	for (int i = 2; i <= n; i++)
		for (int j = 1; j < i; j++)
			C[i][j] = C[i-1][j-1]+C[i-1][j];
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j < a[i]; j++)
			if (b[j] > 0) {
				LL s = 1;
				b[j]--;
				for (int k = 0, p = n-i; k < 10; p -= b[k++])
					s *= C[p][b[k]];
				b[j]++;
				ans += s;
			}
		b[a[i]]--;
	}
	write(ans);
	return 0;
}