Topic link 1001 1003 1009 The inverse element part of the code in
this article is borrowed from this blog. Link Problem-solving idea: The students who are playing together make it, post it here first, and then study it when you have time. The idea of 1001 is roughly the derivation of geometry , and then find the expectation, which is the sum of its reciprocal from 1 to n divided by n and then multiplied by 3. 1001AC code:1/h^2=1/a^2+1/b^2+1/c^2
#include<iostream>
#include<cstdio>
using namespace std;
long long res[6000005];
long long mod=998244353;
void exgcd(long long a,long long b,long long &gcd,long long &x,long long &y)
{
if(b==0)
{
x=1;
y=0;
gcd=a;
}
else
{
exgcd(b,a%b,gcd,y,x);
y-=x*(a/b);
}
}
long long inv(long long a,long long mod)
{
long long x,y,gcd;
exgcd(a,mod,gcd,x,y);
return gcd==1?(x%mod+mod)%mod:-1;
}
int main()
{
long long T,n;
for(int i=1;i<=6000005;i++)
{
long long j=i;
res[i]=inv(j*j,mod);
}
for(int i=2;i<=6000005;i++)
{
res[i]=(res[i]+res[i-1])%mod;
}
for(int i=1;i<=6000005;i++)
{
res[i]=3*res[i]*inv(i,mod)%mod;
}
scanf("%lld",&T);
while(T--)
{
scanf("%lld",&n);
printf("%lld\n",res[n]);
}
return 0;
}
Both 1003 and 1009 are looking for patterns. . . . . .
1003AC code:
#include<iostream>
#include<cstdio>
#include<set>
#include<algorithm>
#include<stdlib.h>
using namespace std;
typedef long long ll;
long long mod=998244353;
struct number
{
int d,x,y;
bool operator <(const number q)
{
if(y==q.y)return x<q.x;
return y<q.y;
}
}p[1000005];
int main()
{
int T,n,k;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&k);
for(int i=1;i<=2*n*(1<<k);i++)
{
scanf("%d",&p[i].d);
p[i].x=1;
p[i].y=i;
}
for(int i=1;i<=k;i++)
{
for(int j=1;j<=2*n*(1<<k);j++)
{
if(p[j].y<=2*n*(1<<k)/(1<<i))
{
p[j].x=1+(1<<(i-1))-p[j].x;
p[j].y=1+2*n*(1<<k)/(1<<i)-p[j].y;
}
else
{
p[j].x+=1<<(i-1);
p[j].y-=2*n*(1<<k)/(1<<i);
}
}
}sort(p+1,p+1+2*n*(1<<k));
for(int i=1;i<=2*n*(1<<k);i++)
{
if(i==1)printf("%d",p[i].d);
else printf(" %d",p[i].d);
}printf("\n");
}
return 0;
}
1009AC code:
#include<iostream>
#include<cstdio>
using namespace std;
typedef long long ll;
//long long res[6000005];
long long mod=998244353;
ll qpow(ll m,ll q)
{
ll ans=1;
while(q)
{
if(q&1)
ans=ans*m%mod;
m=m*m%mod;
q>>=1;
}
return ans;
}
//ll inv[6000005];
ll getinv(ll n)
{
ll num3=qpow(n,mod-2);
return num3;
}
int main()
{
long long T,n;
scanf("%lld",&T);
while(T--)
{
scanf("%lld",&n);
ll res=(qpow(2,n)%mod+1%mod+(2%mod)*qpow((3*getinv(2))%mod,n)%mod)%mod;
printf("%lld\n",res);
}
return 0;
}
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