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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#197513 | #6417. Classical Summation Problem | ucup-team870# | WA | 11ms | 19212kb | C++14 | 1.0kb | 2023-10-02 16:36:51 | 2023-10-02 16:36:52 |
Judging History
answer
#include <bits/stdc++.h>
#define rep(i,l,r) for(int i=l; i<=r; i++)
#define per(i,r,l) for(int i=r; i>=l; i--)
#define IOS {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);}
using namespace std;
typedef long long ll;
const int N=1e6+6,mod=998244353;
void add(ll &x,ll y){x=(x+y)%mod;}
ll qp(ll x,ll y){
ll res=1;
while(y){
if(y&1)res=res*x%mod;
x=x*x%mod; y>>=1;
}
return res;
}
ll fac[N],inv[N];
ll C(int i,int j){
return fac[i]*inv[j]%mod*inv[i-j]%mod;
}
int main() {
const int M=1e6;
fac[0]=1; rep(i,1,M)fac[i]=fac[i-1]*i%mod;
inv[M]=qp(fac[M],mod-2); per(i,M,1)inv[i-1]=inv[i]*i%mod;
int n,k;cin>>n>>k;
ll ans=0,nk=qp(n,k);
rep(i,1,n/2){
if(k&1){
add(ans,nk);
}
else{
add(ans,nk+C(k,k/2)*qp((1ll*i*(n-i)%mod),k/2) );
}
}
if(n%2==0){
int x=n/2; ll v=qp(x,k);
rep(i,(k+1)/2,k)add(ans,C(k,i)*v%mod);
}
cout<<(n*nk-ans+mod)%mod;
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 11ms
memory: 19004kb
input:
3 2
output:
14
result:
ok 1 number(s): "14"
Test #2:
score: 0
Accepted
time: 11ms
memory: 19212kb
input:
5 3
output:
375
result:
ok 1 number(s): "375"
Test #3:
score: -100
Wrong Answer
time: 7ms
memory: 19044kb
input:
2 2
output:
998244352
result:
wrong answer 1st numbers differ - expected: '5', found: '998244352'