Codeforces Round #439 (Div. 2) C. The Intriguing Obsession 组合数学

传送门

题意：（太懒了）（逃）

做法：

对于蓝色岛屿群和红色岛屿群，现蓝色和红色之间有k座桥，那么对于蓝色岛屿总数假设为A，那么要在A座桥里选k座桥，对于红色岛屿总数假设为B，在B座桥里选k座桥，

那么就是C(A,k)*C(B,k) 可是这些桥之间如果顺序不同也算做不同的方案，那么也就是k!，所以蓝色和红色就是C(A,k)*C(B,k)*k!，求和得(懒得写markdown了 逃)sigma_(k=0)到(min(A,B)) 也就是k的选择在0到min(A,B)之间，那么对于B,C同理，A,C同理，乘出来就是答案啦。

//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)#define S_1(x) scan_d(x)#define S_2(x,y) scan_d(x),scan_d(y)#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef long long LL;typedef pair <int, int> ii;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=4e3+10;const int maxx=4e5+10;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;const int MOD=998244353;const int N=5e4+5;namespace COMB{    LL F[N<<1], Finv[N<<1], inv[N<<1];    void init()    {        inv[1] = 1;        for (int i = 2; i < N<<1; i++)        {            inv[i] = (LL)(MOD - MOD/i) * inv[MOD%i] % MOD;        }        F[0] = Finv[0] = 1;        for (int i = 1; i < N<<1; i++)        {            F[i] = (LL)F[i-1] * i % MOD;            Finv[i] = (LL)Finv[i-1] * inv[i] % MOD;        }    }    int comb(int n, int m)    {        if (m < 0 || m > n) return 0;        return (LL)F[n] * Finv[n-m] % MOD * Finv[m] % MOD;    }}using namespace COMB;void solve(){    int a,b,c;    W(s_3(a,b,c)!=EOF)    {        LL ans1=0,ans2=0,ans3=0;        FOR(0,min(a,b),i)            ans1=(ans1+F[i]%MOD*comb(a,i)%MOD*comb(b,i)%MOD)%MOD;        FOR(0,min(a,c),i)            ans2=(ans2+F[i]%MOD*comb(a,i)%MOD*comb(c,i)%MOD)%MOD;        FOR(0,min(b,c),i)            ans3=(ans3+F[i]%MOD*comb(b,i)%MOD*comb(c,i)%MOD)%MOD;        print((ans1*ans2%MOD*ans3%MOD+MOD)%MOD);    }}int main(){    //freopen( "in.txt" , "r" , stdin );    //freopen( "data.txt" , "w" , stdout );    int t=1;    //s_1(t);    init();    for(int cas=1;cas<=t;cas++)    {        //printf("Case %d:\n",cas);        solve();    }}