题解 msc的背包

传送门

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【分析】

不难写出，生成函数 (displaystyle F(x)=({1over 1-x})^ncdot ({1over 1-x^2})^m={(1+x)^nover (1-x^2)^{n+m}})

故 (displaystyle F[k]=sum_{i+j=k} [2mid j]dbinom{n}{i}dbinom{-(n+m)}{jover 2}=sum_{i=0}^n[2mid k-i]dbinom{n}{i} dbinom{n+m-1+{k-iover 2}}{n+m-1})

考虑 (displaystyle dbinom n i={(n-i)cdot (n-i+1)over (i+1)cdot (i+2)}dbinom n {i-2})

且 (displaystyle dbinom{n+m-1+j}{n+m-1}={(n+m-1+j)^{underline{n+m-1}}over (n+m-1)!}=dbinom {n+m-1+j}{j}={n+m+j-1over j}cdot dbinom{n+m-1+(j-i)}{j-1})

记 (res_1=dbinom{n}{kmod 2}, res_2=dbinom{n+m+(k/2)-1}{n+m-1})

而后 (F[k]+=res_1cdot res_2)

新的由 (displaystyle res_1={(n-i)cdot (n-i-1)over icdot (i+1)}cdot res_1, res_2={jover n+m+j-1}cdot res_2) 可 (O(log n)) 生成

需要考虑快速幂的复杂度

总复杂度转为遍历复杂度，即 (O(nlog n))

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【代码】

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pii;
typedef double db;
#define fi first
#define se second
#define int ll
const int MOD=998244353;
ll n, m, k;
inline ll fpow(ll a,ll x) { ll ans=1; for(;x;x>>=1,a=a*a%MOD) if(x&1) ans=ans*a%MOD; return ans; }
inline ll ans(){
    ll res1, res2;
    res1=1, res2=1;
    for(int i=1, j=n+m+(k>>1)-1, I=n+m-1; i<=I; ++i, --j)
        res1=res1*j%MOD, res2=res2*i%MOD;
    res2=res1*fpow(res2, MOD-2)%MOD;
    res1=((k&1)?n:1);

    ll res=0;
    for(int i=k&1, j=k>>1;i<=n&&j>=0;i+=2, --j){
        res = (res+res1*res2)%MOD;
        res1 = res1*(n-i)%MOD*(n-i-1)%MOD*fpow( (i+2)*(i+1)%MOD, MOD-2 )%MOD;
        res2 = res2*j%MOD*fpow(n+m+j-1, MOD-2)%MOD;
    }
    return res;
}
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    cin>>n>>m>>k;
    cout<<ans();
    cout.flush();
    return 0;
}