Codeforces Round 875 (Div. 2)(D)

合集 - acm(93)

1.AtCoder Beginner Contest 293(C,D ,E,F)2023-03-172.Educational Codeforces Round 115 (Rated for Div. 2)（D,E)2023-03-183.AtCoder Beginner Contest 294(E,F,G)2023-04-034.AtCoder Beginner Contest 2462023-03-295.中国石油大学（北京）第三届“骏码杯”程序设计竞赛（同步赛）(D,E,F)2023-03-256.Codeforces Global Round 16(D,E,F)2023-03-257.AtCoder Beginner Contest 209（D，E）2023-05-088.Monoxer Programming Contest 2022（AtCoder Beginner Contest 238）（E，F）2023-05-079.AtCoder Beginner Contest 285（B,D,E,F)2023-05-0610.AtCoder Beginner Contest 242（D，E)2023-05-0311.AtCoder Beginner Contest 223（D,E,F)2023-04-1512.AtCoder Beginner Contest 207（D,E)2023-04-1113.AtCoder Beginner Contest 247（E,F)2023-04-1014.AtCoder Beginner Contest 226(E,F,G)2023-04-0515.AtCoder Beginner Contest 229(F,G)2023-06-2316.AtCoder Beginner Contest 273(E)2023-06-1317.AtCoder Beginner Contest 286(G)2023-06-0218.AtCoder Beginner Contest 287(C,D,E,F)2023-06-0119.AtCoder Beginner Contest 288(D,E,F)2023-05-3120.AtCoder Beginner Contest 289(E,F)2023-05-3021.AtCoder Beginner Contest 290(D,E)2023-05-2922.AtCoder Beginner Contest 292(E,F,G)2023-05-2823.AtCoder Beginner Contest 298(D,F)2023-05-2724.AtCoder Beginner Contest 299(E,F)2023-05-2725.AtCoder Beginner Contest 300(E,F)2023-05-2526.AtCoder Beginner Contest 302（E，F，G）2023-05-2427.AtCoder Beginner Contest 253(E,F)2023-05-1728.AtCoder Beginner Contest 245(D，E，F)2023-05-1529.AtCoder Beginner Contest 248(D，E，F) 2023-05-1430.AtCoder Beginner Contest 206（Sponsored by Panasonic）（E，F）2023-05-09

31.Codeforces Round 875 (Div. 2)(D)2023-07-08

32.AtCoder Beginner Contest 178(E,F)2023-07-0833.AtCoder Beginner Contest 307（E,F,G)2023-07-0134.CodeTON Round 5 (Div. 1 + Div. 2, Rated, Prizes!)C2023-06-3035.Educational Codeforces Round 151 (Rated for Div. 2)(C,D)2023-06-3036.AtCoder Beginner Contest 212(E,F)2023-06-2437.牛客小白月赛512022-12-0938.2022年浙大城市学院新生程序设计竞赛（同步赛）（补题）2022-12-1139.Codeforces Round #770 (Div. 2)B，C2022-12-2440.Codeforces Global Round 24(B,C)2022-12-2341.Codeforces Round #836 (Div. 2）C2022-12-2142.Codeforces Round #840 (Div. 2) C2022-12-2043.Good Bye 2022: 2023 is NEAR C2022-12-3144.Codeforces Round #765 (Div. 2)A，B，C2022-12-3145.Codeforces Round #766 (Div. 2)C，D2022-12-2946.Codeforces Round #841 (Div. 2) and Divide by Zero 20222022-12-2847.Codeforces Round #767 (Div. 2)C ,D 2022-12-2748.Codeforces Round #768 (Div. 2)C ,D2022-12-2649. Codeforces Round #769 (Div. 2) B，C2022-12-2550.Educational Codeforces Round 122 (Rated for Div. 2)，C，D2022-12-2551.Educational Codeforces Round 119 (Rated for Div. 2)2023-01-1452.AtCoder Beginner Contest 2582023-01-1353.Codeforces Round #763 (Div. 2)C2023-01-1254.Codeforces Round #843 (Div. 2)（B，C，D，E）2023-01-1255.Educational Codeforces Round 141 (Rated for Div. 2)（B，C，D）2023-01-1056.AtCoder Beginner Contest 275（B，C，D，E，F）2023-01-0957.Codeforces Round #842 (Div. 2)（B，D，E）2023-01-0958.AtCoder Beginner Contest 284（D，E，F）2023-01-0859.The 14th Jilin Provincial Collegiate Programming Contest（补题）2023-01-0760.牛客小白月赛65（C，D，E，F）2023-01-0761.AtCoder Beginner Contest 281（D，E，F）2023-01-0562.Good Bye 2021: 2022 is NEAR D2023-01-0563.Hello 20232023-01-0464.The 15th Jilin Provincial Collegiate Programming Contest(补题)2023-01-0365.Codeforces Round #781 (Div. 2)C 2023-01-0266.Hello 2022（B，D）2023-01-0167.AtCoder Beginner Contest 272（D,E)2023-03-1068.Codeforces Round 751 (Div. 2)（D）2023-03-0869.Codeforces Round 856 (Div. 2)(C,D)2023-03-0870.Codeforces Round 752 (Div. 2)（C，D，E）2023-03-0871.Codeforces Round 855 (Div. 3)（E，F）2023-03-0572.AtCoder Regular Contest 131（A,B,C）2023-03-0573.Educational Codeforces Round 144 (Rated for Div. 2)（A，B，C，D）2023-03-0574.Codeforces Round 853 (Div. 2)（C,D)2023-03-0375.牛客练习赛109（C,D)2023-03-0376.AtCoder Beginner Contest 291（Sponsored by TOYOTA SYSTEMS）（D，E，F）2023-03-0177.Educational Codeforces Round 143 (Rated for Div. 2)（A，C,D)2023-02-2878.Codeforces Round #852 (Div. 2)（C,D)2023-02-1379.Educational Codeforces Round 118 (Rated for Div. 2)（D，E）2023-02-0980.AtCoder Beginner Contest 236（D，E，F）2023-02-0981.Codeforces Round #850 (Div. 2, based on VK Cup 2022 - Final Round)(B,D)2023-02-0882.Codeforces Round #848 (Div. 2)（B,C,D）2023-02-0783.TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) (B,C,D) 2023-02-0784.Codeforces Round #846 (Div. 2)（B，E） 2023-02-0785.Educational Codeforces Round 142 (Rated for Div. 2)（C，D）2023-02-0486.2023牛客寒假算法基础集训营62023-02-0487.Codeforces Round #845 (Div. 2) and ByteRace 2023（A，B，C）2023-02-0288.2023牛客寒假算法基础集训营5 2023-02-0289.2023牛客寒假算法基础集训营3 2023-01-2190.2023牛客寒假算法基础集训营22023-01-1991.2023牛客寒假算法基础集训营12023-01-1892.Educational Codeforces Round 120 (Rated for Div. 2) C，D2023-01-1593.AtCoder Beginner Contest 254（C，D，E，F) 2023-01-15

收起

Codeforces Round 875 (Div. 2)(D)

D （思维）

这个题意是给你两个数组，$a$和$b$,我们需要找到这样的二元组$(i,j)$满足$a_i\times a_j=b_i+b_j$,问一共有多少组满足以上条件的二元组

题目还告诉我们数组里面的数字都是不大于$n$的，那么因为两个数相乘的范围一定是$1-n$的，那么我们可以枚举出较小的那一个$a$，然后再对其他的$n$个$a$和$b$,那么已知三项，可以求出剩余的一项，就是和我们枚举的这个$a$所对应的$b$,所以，我们也要记录当$a$为此时枚举的数字是对应的$b$的数量，可以用一个$vector$来存（用$map$可能会超时，正好数据范围也不大）

但是还有一个小细节的地方，就是我们需要保证$a$是有序的（这个解法是在知乎博主那里看到的）而且必须是从大到小的

题解

我后面自己试验了一下，如果不是有序的，那么最后获得的答案可能会少

假设此时的$a1$是$1$

后面有$(a,b)(1,1),(1,3)(3,2)(6,3)$只有按照从小到大的顺序才可，其他的都是不对的

其中原因我也不是很清楚

这样的话，我们可以造成二元组的一定是$a$大于等于$a1$，$a$的乘积越来越大，对于同一个$a$,$b$是递增的，那么我们所需的$a2$是在递减，但是感觉关系不大，还不是很懂为什么一定要排序

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include<cmath>
#include <unordered_map>
#include <array>
#include <cstring>
#include <bitset>
#include <numeric>
using namespace std;
#define int long long 
#define LL long long
#define ios  ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define inf 1e18
#define INF 1e18
#define eps 1e-10
#define mem(a,b) memset((a),(b),sizeof(a))
const int maxn=1e6+10;
const int mod=1e9+7;
int tt;
int n;
pair<int,int>x[maxn];
void solve()
{
  cin>>n;
  for(int i=1;i<=n;i++)
  {
    cin>>x[i].first;
  }
  for(int i=1;i<=n;i++)
  {
    cin>>x[i].second;
  }
  sort(x+1,x+1+n);
  int ans=0;
  for(int i=1;i*i<=n<<1;i++)
  {
    vector<int>cnt(n+1,0);
    int a1=i;
    for(int j=1;j<=n;j++)
    {
      int a2=x[j].first;
      int b2=x[j].second;
      int b1=a1*a2-b2;
      if(b1>=1&&b1<=n)
      {
        ans+=cnt[b1];
      }
      if(a1==a2)
      {
        cnt[b2]++;
      }
    }
  }
  cout<<ans<<"\n";
  return ;
}
signed main ()
{
  ios;
 // cin>>t;
  tt=1;
  cin>>tt;
  while (tt--)
  {
    solve();
  }
  system ("pause");
  return 0;
}