Wannafly挑战赛17 B

Posted 2020-11-14 sssy

题解

大概就是求证这个
[sum_i^nC_{n}^i*C_n^i = C_{2n}^n]
证明：
[(1+x)^{2n} = [C(0,n)+C(1,n)*x+...+C(n,n)*x^n]*[[C(0,n)+C(1,n)*x+...+C(n,n)*x^n]]]
[=...+[C(0,n)*C(n,n)+C(n-1,n)+...+C(n,n)*C(0,n)]x^n+...]
也就是说，在((1+x)^{2n})的展开式中，(x^n)的系数是
[sum_k^nC(k,n)*C(n-k) = sum_k^nC(k,n)^2]
根据二项式定理
[(1+x)^{2n}=sum_k^{2n}[C(k,2n)]*x^k]即(x^k)的系数为C(n,2n),由此可得(sum_k^nC(k,n)^2 = C(n,2n))

代码

#include<cstdio>
#include<algorithm>
#define LL long long
#define mod 998244353
 LL inv(LL x,int y) {
    LL ret = 1;
    for(;y;y >>= 1,x = x * x % mod)
           if(y & 1) ret = ret * x % mod;
    return ret;    
}
const int maxn = 1000007;
LL jc[maxn * 2];
LL C(int a,int b)  {
    return ((((jc[a] * inv(jc[b],mod - 2)% mod) + mod) % mod)
        * inv(jc[a - b],mod - 2) + mod) % mod;
}
int main() {
    jc[0] = jc[1] = 1;
    int n;
    scanf("%d",&n);
    for(int i = 2;i <= 2 * n;++ i)
        jc[i] = jc[i - 1] * i % mod;
    LL ans = 0;
    for(int i = 1;i <= n;++ i)
        ans = (ans + C(2 * i,i)) % mod;
    printf("%lld
",ans);
    return 0;
}