CodeChef STFM - Chef and Strange Formula (康托展开)
思路
康托展开$X=a[n](n-1)!+a[n-1](n-2)!+…+a[i](i-1)!+…+a[1]0!$
那么对照一下题目里的公式。
$F(x) = 11!+ 22!+ 33!+ … + xx! + (1+2+3..+x) * x$
后面的和大家都会算。
前面一时找不到规律，后来请教了hcbbt巨巨。他一眼就看出是康托展开！
前面就是序列
x+1, x, x-1, …, 2, 1 的康托展开。
显然x+1个元素的排列为$(x+1)!$，因为是从0开始计算，所以上面这个排列的康托展开就是$(x+1)!-1$
所以$F(x) = (x+1)!- 1 + x* \frac {x+1}{2}$
之前算的时候不知道大阶乘的模怎么算。今天看别人代码才想到如果阶乘大于模那么值就是0。
真的怀疑当时自己脑子抽了。
代码
#include <stack>
#include <cstdio>
#include <list>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 1e7 + 10;
const int MOD = 1e9 + 7;
const int dir[][2] = { {-1, 0}, {0, -1}, { 1, 0 }, { 0, 1 } };
const int hash_size = 4e5 + 10;
int cases = 0;
typedef pair<int, int> pii;
LL Fac[MAXN], m;
inline LL Mod(LL n)
{
return n % m;
}
int main()
{
//ROP;
LL n;
cin >> n >> m;
Fac[0] = 1;
for (int i = 1; i <= m; i++) Fac[i] = Mod(Fac[i-1] * i);
LL ans = 0;
for (int i = 0; i < n; i++)
{
//先计算sum
LL tmp, tmpAns = 0;
cin >> tmp;
if (tmp & 1) tmpAns = Mod(Mod(tmp) * Mod(Mod(tmp) * Mod((tmp+1)/2)));
else tmpAns = Mod(Mod(tmp) * Mod(Mod(tmp>>1) * (Mod(tmp+1))));
//compute factorial
LL fac;
if (tmp >= m-1) fac = 0;
else fac = Fac[tmp+1];
tmpAns = Mod(tmpAns + fac - 1);
ans = Mod(ans + tmpAns);
}
cout << ans << endl;
return 0;
}