求助 CF D 为什么会 T
  • 板块学术版
  • 楼主UperFicial
  • 当前回复7
  • 已保存回复7
  • 发布时间2023/5/29 01:58
  • 上次更新2023/10/23 14:24:22
查看原帖
求助 CF D 为什么会 T
360511
UperFicial楼主2023/5/29 01:58
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<cstring>
#include<cstdlib>
#include<map>
#include<ctime>
#include<assert.h>
#include<unordered_map>
#define rep(i,a,b) for(register int i=a;i<=b;++i)
#define rev(i,a,b) for(register int i=a;i>=b;--i)
#define gra(i,u) for(register int i=head[u];i;i=edge[i].nxt)
#define Clear(a) memset(a,0,sizeof(a))
#define yes puts("YES")
#define no puts("NO")
using namespace std;
typedef long long ll;
const int INF(1e9+10);
const ll LLINF(1e18+10);
inline int read()
{
	int s=0,w=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')s=s*10+(ch-'0'),ch=getchar();
	return s*w;
}
template<typename T>
inline T Min(T x,T y){return x<y?x:y;}
template<typename T>
inline T Max(T x,T y){return x>y?x:y;}
template<typename T>
inline void Swap(T&x,T&y){T t=x;x=y;y=t;return;}
template<typename T>
inline T Abs(T x){return x<0?-x:x;}

const int MOD(1e9+7);
template<typename T>
inline T add(T x){return x;}
template<typename T,typename... types>
inline T add(T x,types... y){T z=add<T>(y...);return x+z>=MOD?x+z-MOD:x+z;}
template<typename T>
inline T mul(T x){return x;}
template<typename T,typename... types>
inline T mul(T x,types... y){return (ll)x*mul<T>(y...)%MOD;}
inline int sub(int x,int y){return add(x-y,MOD);}

const int MAXN(2e5+10);

int n,a[MAXN];
int t;
vector<int>G[MAXN]; 
map<int,int>mp[MAXN];
ll ans;

int main()
{
//	freopen("read.txt","r",stdin);
	t=read();
//	assert(t==1);
	while(t--)
	{
		ans=0;
		n=read();
//		assert(n<200000);
		rep(i,1,n) a[i]=read();
		rep(i,1,n)
		{
			int x=read();
			G[a[i]].push_back(x),mp[a[i]][x]++;
			if(1ll*a[i]*a[i]==x+x) --ans;
		}
		rep(x,1,n)
		{
			for(auto y:G[x])
			{
				if(mp[1].count(x-y)!=0) ans=ans+mp[1][x-y];
				int bj=x-y;
				rep(i,2,n)
				{
					bj+=x;
					if(bj>n) break;
					if(mp[i].count(bj)!=0) ans=ans+mp[i][bj];
				}
			}
		}
		printf("%lld\n",ans/2);
		rep(i,1,n) G[i].clear(),mp[i].clear();
	}
	return 0;
}
2023/5/29 01:58
加载中...