Codeforces Round #491 (Div. 2) E Bus Number —— 排列组合

题目链接:http://codeforces.com/contest/991/problem/E
有重复问题的全排列是 A n n A n u m [ 0 ] n u m [ 0 ] A n u m [ 1 ] n u m [ 1 ] . . . A n u m [ 9 ] n u m [ 9 ]
然后我们只要减去开头为0的全拍列就好了。
dfs是搜索每个数字有多少个

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll fac[25];
void init()
{
    fac[1]=fac[0]=1;
    for(ll i=2;i<20;i++)
        fac[i]=i*fac[i-1];
}
int num[15];
ll ans;
int a[15];
void dfs(int x)
{
    if(x>9)
    {
        ll mul=1;
        int sum=0;
        for(int i=0;i<=9;i++)
            mul*=fac[a[i]],sum+=a[i];
        mul=fac[sum]/mul;
        ll dec=1;
        if(a[0])
        {
            dec=fac[a[0]-1];
            for(int i=1;i<=9;i++)
                dec*=fac[a[i]];
            dec=fac[sum-1]/dec;
            mul-=dec;
        }
        ans+=mul;
        return ;
    }
    for(int i=1;i<=num[x];i++)
    {
        a[x]=i;
        dfs(x+1);
    }
    if(!num[x])
        dfs(x+1);
}
int main()
{
    init();
    ll n;
    scanf("%lld",&n);
    int len=0;
    while(n)
    {
        num[n%10]++;
        n/=10;
    }
    dfs(0);
    printf("%lld\n",ans);
    return 0;
}

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转载自blog.csdn.net/tianyizhicheng/article/details/82659748