👊E👊题👊要👊取👊模👊
2026-08-15 22:05:15
发布于:广东
ABC E.
板子。拆贡献。 独立考虑,在剩下 个数中选 个数的所有方案都可以贡献一次。然后是式子的乘积贡献,求出每个元素在剩下 中选出 个数中的贡献,进行求和,可以快速得出与其他元素的乘积之和。
为什么,最后要,再模一次,我不是,已经,模了,无数次吗,最后再模,是用来,干什么的,防复活?
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mn(x,y) ((x<y)?(x):(y))
#define mx(x,y) ((x>y)?(x):(y))
#define mid (l+r>>1)
#define int long long
ll n,k,a[200005],b[200005],t[200005],mod=998244353;ll ans,sum,ans2;
ll inv(ll x){
ll p=mod-2,ans=1;
while(p){
if(p&1)ans=ans*x%mod;
x=x*x%mod;
p>>=1;
}
return ans;
}
inline long long read(){
long long x=0,f=1,ch=getchar_unlocked();
for(;!isdigit(ch);ch=getchar_unlocked())if(ch=='-')f=-1;
for(;isdigit(ch);ch=getchar_unlocked())x=(x<<3)+(x<<1)+(ch^48);
return x*f;
}
inline void write(long long x){
x<0?x=-x,putchar_unlocked('-'):0;static short st[19],top(0);
do st[++top]=x%10,x/=10;while(x);
while(top)putchar_unlocked(st[top--]|48);
}
signed main(){
n=read(),k=read();
for(int i=1;i<=n;++i)a[i]=read();
t[0]=1;
for(int i=1;i<=n;++i)t[i]=t[i-1]*i%mod;
ll s=(((t[n-2]*inv(t[k-2]))%mod)*inv(t[n-k])%mod);
for(int i=1;i<=n;++i){
b[i]=s*a[i]%mod;
sum=(sum+b[i])%mod;
}
ll ae=(t[n-1]*inv(t[k-1])%mod)*inv(t[n-k])%mod;
for(int i=1;i<=n;++i){
ans=(ans+((((ae*a[i]%mod)%mod)*a[i])%mod))%mod+(((sum+mod-b[i])%mod)*a[i]%mod)%mod;
}
write(ans%mod);
return 0;
}
我真的,求,你,了!AtC,你给我,听,好了,我,背后的,势力,超你。
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
👊E👊题👊要👊取👊模👊
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