P2524 Uim的情人节礼物·其之弐【康托展开模板题】

题目链接

  我在这里加了树状数组来优化康托展开，但是这道题的数据其实很小，不需要加也是可以的。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 1e2 + 7;
int trie[maxN], N;
inline void add(int i)
{
    while(i <= N)
    {
        trie[i]++;
        i += lowbit(i);
    }
}
inline ll query(int x)
{
    ll ans = 0;
    while(x)
    {
        ans += trie[x];
        x -= lowbit(x);
    }
    return ans;
}
ll jc[maxN];
ll Cantor(int len, char *a)
{
    ll tmp, ans = 0;
    for(int i=len-1; i>=0; i--)
    {
        tmp = query(a[i] - '0');
        ans += tmp * jc[len - i - 1];
        add(a[i] - '0');
    }
    return ans + 1;
}
char s[maxN];
int main()
{
    jc[0] = jc[1] = 1;
    for(ll i=2; i<10; i++) jc[i] = jc[i-1] * i;
    scanf("%d", &N);
    scanf("%s", s);
    printf("%lld\n", Cantor(N, s));
    return 0;
}