双元素非递增(容斥)--Number Of Permutations Educational Codeforces Round 71 (Rated for Div. 2)
题意:https://codeforc.es/contest/1207/problem/D
n个元素,每个元素有a、b两个属性,问你n个元素的a序列和b序列有多少种排序方法使他们不同时非递减(不同时good)。
思路:
真难则反+容斥,反向考虑,ans1=如果a序列非递减则有a中各个数字出现次数的阶乘的乘积个,ans2=b序列也是一样。
ans3=然后还要减去a序列和b序列都是good的方案数,就是元素相同的出现次数阶乘的乘积(注意,如果不存在双good就不算ans3)。
ANS就是:全排列 - ans1 - ans2 + ans3
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\草稿.txt","r",stdin); 6 #include <bitset> 7 #include <map> 8 //#include<unordered_map> 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr 13 #include <string> 14 #include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor 21 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) 22 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation 23 #define fo(a,b,c) for(register long long a=b;a<=c;++a) 24 #define fr(a,b,c) for(register int a=b;a>=c;--a) 25 #define mem(a,b) memset(a,b,sizeof(a)) 26 #define pr printf 27 #define sc scanf 28 #define ls rt<<1 29 #define rs rt<<1|1 30 typedef long long ll; 31 void swapp(int &a,int &b); 32 double fabss(double a); 33 int maxx(int a,int b); 34 int minn(int a,int b); 35 int Del_bit_1(int n); 36 int lowbit(int n); 37 int abss(int a); 38 //const long long INF=(1LL<<60); 39 const double E=2.718281828; 40 const double PI=acos(-1.0); 41 const int inf=(1<<30); 42 const double ESP=1e-9; 43 const int mod=(int)998244353; 44 const int N=(int)1e6+10; 45 46 ll a[N],b[N]; 47 pair<ll,ll>s[N]; 48 map<pair<ll,ll>,ll>mp3; 49 map<ll,ll> mp1,mp2; 50 51 ll v(ll x) 52 { 53 ll sum=1; 54 fo(i,1,x) 55 sum*=i,sum%=mod; 56 return sum; 57 } 58 59 int main() 60 { 61 int n; 62 ll ans=1,temp1,temp2,temp3; 63 sc("%d",&n); 64 fo(i,1,n)ans*=i,ans%=mod,sc("%lld%lld",&a[i],&b[i]),mp1[a[i]]++,mp2[b[i]]++,s[i]={a[i],b[i]},mp3[s[i]]++; 65 temp1=temp2=temp3=1; 66 for(auto i:mp1) 67 temp1*=v(i.second),temp1%=mod; 68 for(auto i:mp2) 69 temp2*=v(i.second),temp2%=mod; 70 for(auto i:mp3) 71 temp3*=v(i.second),temp3%=mod; 72 sort(s+1,s+1+n); 73 for(int i=1;i<n;++i) 74 if(s[i].second>s[i+1].second) 75 temp3=0; 76 pr("%lld\n",((ans-temp1-temp2+temp3)%mod+mod)%mod); 77 return 0; 78 } 79 80 /**************************************************************************************/ 81 82 int maxx(int a,int b) 83 { 84 return a>b?a:b; 85 } 86 87 void swapp(int &a,int &b) 88 { 89 a^=b^=a^=b; 90 } 91 92 int lowbit(int n) 93 { 94 return n&(-n); 95 } 96 97 int Del_bit_1(int n) 98 { 99 return n&(n-1); 100 } 101 102 int abss(int a) 103 { 104 return a>0?a:-a; 105 } 106 107 double fabss(double a) 108 { 109 return a>0?a:-a; 110 } 111 112 int minn(int a,int b) 113 { 114 return a<b?a:b; 115 }