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

( holds for all xx.
Given two integers nn and mm,
you need to calculate the coefficient of x^mx​m​​ in f(x)f(x),
modulo 998244353998244353.

Input Format

Multiple test cases (no more than 100100).

Each test case contains one line consisting of two integers nn and mm.

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 0
2 1
2 2

样例输出

998244352
0
2

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_back
typedef 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);
}
}
}
}