题意：给定一个长度为n的数组，以及一个整数m，数组中的数ai<=m，要求判断a1!+a2!+a3!+...+an!是否能够整除m!。
思路：千万不要被这题是div2的d题，误判它的难度，从而禁锢了思维，具体的说，任何题目都应该找它的解而不在意解过程的难易程度。对于这题来说，根据阶乘的性质i*(i-1)!==i!，将乘法转换为龟速乘，很明显能够看出，该题是一个记录ai的出现次数，从小向大递推的过程，要求保证i<m,heap[i]==0,i==m,heap[i]!=0
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<string>
#include<bitset>
#include<cmath>
#include<array>
#include<atomic>
#include<sstream>
#include<stack>
#include<iomanip>
//#include<bits/stdc++.h>//#define int ll
#define IOS std::ios::sync_with_stdio(false);std::cin.tie(0);
#define pb push_back
#define endl '\n'
#define x first
#define y second
#define Endl endl
#define pre(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,b,a) for(int i=b;i>=a;i--)
#define si(x) scanf("%d", &x);
#define sl(x) scanf("%lld", &x);
#define ss(x) scanf("%s", x);
#define YES {puts("YES");return;}
#define NO {puts("NO"); return;}
#define all(x) x.begin(),x.end()using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<int, PII> PIII;
typedef pair<char, int> PCI;
typedef pair<int, char> PIC;
typedef pair<double, double> PDD;
typedef pair<ll, ll> PLL;
const int N = 500010, M = 2 * N, B = N, MOD = 998244353;
const double eps = 1e-7;
const int INF = 0x3f3f3f3f;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;//int dx[4] = { -1,0,1,0 }, dy[4] = { 0,1,0,-1 };
int dx[8] = { 1,2,2,1,-1,-2,-2,-1 }, dy[8] = { 2,1,-1,-2,-2,-1,1,2 };
int n, m, k;
int a[N];
int prime[N], countNum;
bool st[N];
int heap[N];ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lowbit(ll x) { return x & -x; }
ll qmi(ll a, ll b, ll MOD) {ll res = 1;while (b) {if (b & 1) res = res * a % MOD;a = a * a % MOD;b >>= 1;}return res;
}inline void init() {for (int i = 2; i < N; i++){if (!st[i])prime[countNum++] = i;for (int j = 0; prime[j] * i < N; j++){st[prime[j] * i] = true;if (i % prime[j] == 0)break;}}
}void slove()
{cin >> n >> m;for (int i = 1; i <= n; i++) cin >> a[i];for (int i = 1; i <= n; i++){heap[a[i]]++;}for (int i = 1; i < m; i++){while (heap[i]>0)heap[i] -= (i + 1), heap[i + 1]++;if(heap[i]!=0)NO}if(heap[m])YESelse NO
}signed main()
{//IOS;int _ = 1;//si(_);init();while (_--){slove();}return 0;
}
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