2017 ACM-ICPC 亚洲区（西安赛区）网络赛 F. Trig Function cos(nx)

( holds for all $x$.

Given two integers $n$ and $m$, you need to calculate the coefficient of x^mx​m​​ in $f(x)$, modulo $998244353$.

Input Format

Multiple test cases (no more than $100$).

Each test case contains one line consisting of two integers $n$ and $m$.

1 \le n \le 10^9,0 \le m \le 10 ^ 41≤n≤10​9​​,0≤m≤10​4​​.

Output Format

Output the answer in a single line for each test case.

样例输入

2 02 12 2

样例输出

99824435202

题目来源

2017 ACM-ICPC 亚洲区（西安赛区）网络赛

就阶乘那里 因为大的一部分可以把小的一部分约掉，然后经过观察得到只有k=0的时候n+2k-2   n-2k的大小是 下面大 上面小，

同理 就特判下k=0就完了。

公式题，记录一下结论

//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 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;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=1e5+10;const int maxx=2e5+100;const double EPS=1e-8;const double eps=1e-8;const int mod=998244353;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;LL fac[maxn];void init(){    fac[0]=1;    for(int i=1;i<=maxn;i++)    {        fac[i]=fac[i-1]*i;        fac[i]%=mod;    }}void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d){    if (!b) {d = a, x = 1, y = 0;}    else{        ex_gcd(b, a % b, y, x, d);        y -= x * (a / b);    }}LL inv2(LL t, LL p){//如果不存在，返回-1    LL d, x, y;    ex_gcd(t, p, x, y, d);    return d == 1 ? (x % p + p) % p : -1;}int n,m;int main(){    init();    W(s_2(n,m)!=EOF)    {        int k,t;        LL fenzi,fenmu;        if(n%2==0)        {            if(m&1)            {                puts("0");            }            else// m偶            {                k=m/2;                t=(n-2*k)/2;                if(t&1)                {                    t=-1;                }                else                {                    t=1;                }                fenmu=fac[2*k];                LL cnt=1;                for(LL i=(LL)n-2*k+2;i<=(LL)(n+2*k-2);i+=2)//上面大                {                    cnt=cnt*i%mod;                }                //cout<<cnt<<endl;                if(k==0)                {                    fenmu=fenmu*n%mod;                }                fenzi=(t*cnt*n%mod+mod)%mod;                fenzi=fenzi*inv2(fenmu,mod)%mod;                print(fenzi);            }        }        else        {            if(m%2==0)            {                puts("0");            }            else //m 奇            {                k=(m+1)/2;                t=(n+1-2*k)/2;                if(t&1)                {                    t=-1;                }                else                {                    t=1;                }                fenmu=fac[2*k-1];                LL cnt=1;                for(LL i=(LL)n+1-2*k+2;i<=(LL)(n+2*k-3);i+=2)                {                    cnt=cnt*i%mod;                }                if(k==0)                {                    fenmu=fenmu*(n-1)%mod*(n+1)%mod;                }                fenzi=(t*cnt*n%mod+mod)%mod;                fenzi=fenzi*inv2(fenmu,mod)%mod;                print(fenzi);            }        }    }}