2017 ACM-ICPC 亚洲区（西安赛区）网络赛 F.Trig Function(论文+组合数)

传送门

f(cos(x))=cos(n∗x) holds for all x.

Given two integers n and m, you need to calculate the coefficient of xm 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≤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

题目大意：

给出f(cos(x))=cos(n∗x) ，求xm前面的系数，其实就是求 cos(x)m 前面的系数。

解题思路：

首先给出论文：传送门
然后就根据论文进行计算就OK了，需要特判两个点。
给出组合数公式：

1) m==0 的时候
2) m==n 的时候

代码：

#include <iostream>#include <string.h>#include <string>#include <algorithm>#include <stdio.h>#include <stdlib.h>#include <math.h>#include <map>using namespace std;typedef long long LL;const int MAXN = 1e4+5;const double PI = acos(-1);const double eps = 1e-8;const LL MOD = 998244353;LL Pow(LL a, LL b){    LL ans = 1;    while(b){        if(b & 1) ans = ans * a % MOD;        b>>=1;        a = a * a % MOD;    }    return ans;}LL Inv[MAXN];void Init(){    Inv[0] = Inv[1] = 1;    for(int i=2; i<MAXN; i++) Inv[i] = (MOD - MOD / i) * Inv[MOD % i] % MOD;    for(int i=2; i<MAXN; i++) Inv[i] = Inv[i]*Inv[i-1]%MOD;}int main(){    //freopen("C:/Users/yaonie/Desktop/in.txt", "r", stdin);    //freopen("C:/Users/yaonie/Desktop/out.txt", "w", stdout);    Init();    LL n, m;    while(~scanf("%lld%lld", &n, &m)){        if(((n&1)&&!(m&1)) || (!(n&1)&&(m&1))){            puts("0");            continue;        }        LL k = (n-m)/2;        if(m == 0){            if(k & 1) puts("998244352");            else puts("1");            continue;        }        if(n == m){            printf("%lld\n",Pow(2LL, n-1));            continue;        }        LL t1 = min(k, n-2*k), t2 = min(k-1, n-2*k);        LL ans1 = Inv[t1], ans2 = Inv[t2];        for(LL i=1; i<=t1; i++){            LL tmp = (n-k-i+1)%MOD;            ans1 = ans1*tmp%MOD;        }        for(LL i=1; i<=t2; i++){            LL tmp = (n-k-i)%MOD;            ans2 = ans2*tmp%MOD;        }        LL ans = (ans1 + ans2) % MOD;        if(k & 1) ans=-ans;        ans = (ans+MOD)%MOD;        ans = ans*Pow(2LL, n-1-2*k)%MOD;        printf("%lld\n",ans);    }    return 0;}