題面

luogu

題解

本來想練數位dp的,結果又忍不住寫了組合數..

去掉一個\(0\)可以看作把\(0\)移到前面去

那麽題目轉化為 \(n\)有多少個排列小於\(n\)

強制某一位比\(n\)的對應位置上的數小, 後面方案組合數算一下即可

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;
}
 

洛谷 P2518 [HAOI2010]計數 (組合數)