2017 ACM-ICPC 亚洲区（西安赛区）网络赛 F

f(cos(x))=cos(n∗x) holds for all xxx.

Given two integers nnn and mmm, you need to calculate the coefficient of xmx^mxm in f(x)f(x)f(x), modulo 998244353998244353998244353.

Input Format

Multiple test cases (no more than 100100100).

Each test case contains one line consisting of two integers nnn and mmm.

1≤n≤109,0≤m≤1041 \le n \le 10^9,0 \le m \le 10 ^ 41≤n≤109,0≤m≤104.

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<cstdlib>
#include<string>
#include<set>
#include<stack>
#define mod 998244353using namespace std;int n, k;long long  PowerMod(long long  a, long long  b = mod - 2, long long c = mod)
{long long  ans = 1;a = a % c;while(b>0){if(b & 1)ans = (ans * a) % c;b >>= 1;a = (a * a) % c;}return ans;
}
void xishu()
{long long ans = 1;if((n & 1) + (k & 1) == 1){cout << 0 << endl;return ;}else{int tmp = (n - k) / 2;if(tmp & 1)ans = -1;elseans = 1;}for(int i = 1; i <= k; i++){ans = ans * PowerMod(i) % mod;}int low = min(n + k - 2, n - k);int high = max(n + k - 2, n - k);if(n + k - 2 <= n - k){for(int i = high ; i > low ; i -= 2){ans = ans * PowerMod(i) % mod;}}else{for(int i = high; i > low; i -= 2){ans = ans * i % mod;}}ans = ans * n % mod;cout << (ans + mod) % mod << endl;
}int main()
{while(~scanf("%d %d", &n, &k)){xishu();}return 0;
}

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