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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
#234842	#6640. Talk That Talk	ugly2333	TL	640ms	29892kb	C++20	1.2kb	2023-11-01 23:42:09	2023-11-01 23:42:09

Judging History

你现在查看的是最新测评结果

[2023-11-01 23:42:09]
评测

测评结果：TL
用时：640ms
内存：29892kb

查看

[2023-11-01 23:42:09]
提交

answer

//Δ_J
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef double DB;
typedef __int128 i128;
const int N = 2222222;
int n,a[N],b[N],c[N];
LL p;
LL fpow(LL x,LL y){
	LL z=1;
	while(y){
		if(y&1)
			z=(i128)z*x%p;
		x=(i128)x*x%p;
		y>>=1;
	}
	return z;
}
LL geta(LL x){
	return fpow(x,(p-1)/2)==1;
}
LL solve(){
	int i;
	LL s=0;
	for(i=1;i<=n*2;i++)
		c[i]=c[i-1]+a[i];
	for(i=1;i<=n;i++)
		if(!a[i])
			s+=c[i+n]-c[i+i];
	for(i=1;i<=n;i++)
		if(!b[i])
			s+=c[n-i];
	c[1]=b[1];
	for(i=2;i<=n;i++)
		c[i]=c[i-2]+b[i];
	for(i=2;i<=n*2;i++)
		if(!a[i])
			s+=c[min(i-2,n*2-i)];
	return s;
}
int main(){
	int T,i;
	LL s;
	scanf("%d",&T);
	while(T--){
		scanf("%lld%d",&p,&n);
		n=min((LL)n,(p-1)/2);
		if(p%4==1){
			s=p-2-(p-1)/2;
			if(geta(2))
				s=(s-1)/2;
			else
				s=(s+1)/2;
			s--;
		}
		else{
			s=(p-1)/2;
			if(geta(2))
				s=(s-1)/2;
			else
				s=(s+1)/2;
			s--;
		}
		s*=n;//cout<<s<<endl;
		for(i=1;i<=n*2;i++)
			a[i]=geta(i);
		for(i=1;i<=n*2;i++)
			b[i]=a[i]^(p%4==3);
		s-=(LL)n*(n-1);
		s+=solve();
		swap(a,b);
		s+=solve();
		printf("%lld\n",s);
	}
	return 0;
}

源文件, 语言: C++20

詳細信息

Test #1:

score: 100

Accepted

time: 640ms

memory: 29892kb

input:

7
7 32
13 1
13 2
67 11
2003 44
1000003 1984
999999999989 987654

output:

0
2
2
146
21510
495014784
246913256130162788

result:

ok 7 numbers

Test #2:

score: -100

Time Limit Exceeded

input:

output: