wa到自闭，终于AC了

AC了 #include<iostream> #include<string.h> using namespace std; int x[1001], y[1001], n; int func(int a, int b) {     int sum = 0;     for (int i = 0; i < n; i++)     {         if (x[i] + y[i] == a || y[i] - x[i] == b)         {             sum++;         }     }     return sum; } int main() {     while (cin >> n)     {         int a[10000] = { 0 };         int b[10000] = { 0 };         memset(x, 0, sizeof(x));         memset(y, 0, sizeof(y));         int max1 = 0, max2 = 0;         int sum = 0;         int ok = 1;         int ak = 1;         for (int i = 0; i < n; i++)         {             cin >> x[i];         }         for (int i = 0; i < n; i++)         {             cin >> y[i];             ok = 1;             ak = 1;             for (int j = 0; j < max1; j++)             {                 if (a[j] == x[i] + y[i])                 {                     ok = 0;                 }             }             if (ok)             {                 a[max1] = x[i] + y[i];                 max1++;             }             for (int j = 0; j < max2; j++)             {                 if (b[j] == y[i] - x[i])                 {                     ak = 0;                 }             }             if (ak)             {                 b[max2] = y[i] - x[i];                 max2++;             }         }         for (int i = 0; i < max1; i++)             {                 for (int j = 0; j < max2; j++)                 {                     if (func(a[i], b[j]) > sum)                     {                         sum = func(a[i], b[j]);                     }                 }             }         cout << sum << endl;     } }

感觉写了一个复杂度好像不太对的做法水过了 #include <bits/stdc++.h> using namespace std; const int MAXN = 1e3 + 5; int n, x[MAXN], y[MAXN]; vector<int> A[MAXN * 2], B[MAXN * 2]; int main() {     scanf("%d", &n);     for (int i = 0; i < n; i++) scanf("%d", &x[i]);     for (int i = 0; i < n; i++) scanf("%d", &y[i]);     for (int i = 0; i < n; i++)     {         A[y[i] + x[i]].push_back(i);         B[y[i] - x[i] + 1000].push_back(i);     }     int ans = 0;     for (int i = 0; i < 2000; i++)     {         for (int j = 0; j < 2000; j++)         {             set<int> s;             for (auto k : A[i]) s.insert(k);             for (auto k : B[j]) s.insert(k);             ans = max(ans, (int)s.size());         }     }     printf("%d\n", ans);     return 0; } /* 4 1 4 4 5 3 0 2 3 */

wa了一个礼拜

赞，ac了。

所谓自闭，原来是这个意思