牛客练习赛68b
牛牛的算术
https://ac.nowcoder.com/acm/contest/7079/B
B,“展开全文”即可正常查看
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
#define fi first
#define se second
#define pk push_back
#define mk make_pair
const ll N=2e5+10, inf=0x3f3f3f3f3f3f3f3f;
ll ijk[N], ans[N], jk[N], kps[N], n;
ll mod=199999;
void init(){
for(ll k=1;k<=mod;k++) kps[k]=(kps[k-1]+k)%mod;
ans[0]=1;
for(ll n=1;n<=mod;n++){
ll i=n;
jk[i]=jk[i-1];
ll j=i;
jk[i]=(jk[i]+j*kps[i]%mod)%mod;
ans[n]=ans[n-1]*i%mod*jk[i]%mod;
}
return ;
}
void work(){
scanf("%lld",&n);
if(n>=mod) puts("0");
else printf("%lld\n",ans[n]);
}
int main() {
init();
ll t;
for(cin>>t;t--;)
work();
return 0;
}
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