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折半枚举#include<iostream> #include<stack> #include<string.h> #include<cmath> #include<iomanip> #include<algorithm> #include<climits> #include<cstdio> #include<vector> #include<sstream> #include<ctype.h> #include<set> #include<map> #include<ctime> #include<stdlib.h> #include<queue> #include<bitset> #define usi unsigned int #define ull unsigned long long using namespace std; template<typename TTT>inline void mr(TTT& theNumberToRead) { theNumberToRead = 0; bool prn = false; char c = getchar(); while (!isdigit(c)) { if (c == '-')prn = true; c = getchar(); } while (isdigit(c)) theNumberToRead = 10 * theNumberToRead + (c ^ 48), c = getchar(); if (prn) theNumberToRead = -theNumberToRead; } template<typename TTT>inline TTT mrr() { TTT theNumberToRead = 0; bool prn = false; char c = getchar(); while (!isdigit(c)) { if (c == '-')prn = true; c = getchar(); } while (isdigit(c)) theNumberToRead = 10 * theNumberToRead + (c ^ 48), c = getchar(); return prn ? -theNumberToRead : theNumberToRead; } template<typename T>void my_write(T x) { if (x) my_write(x / 10), putchar(x % 10 ^ 48); } template<typename T>void mw(T x, char mid) { if (x) { if (x < 0) putchar('-'), x = -x; my_write(x); } else putchar(48); if (mid) putchar(mid); } // ******************************************华丽的分割线****************************************** // ******************************************华丽的分割线****************************************** // ******************************************华丽的分割线****************************************** // ******************************************华丽的分割线****************************************** // ******************************************华丽的分割线****************************************** const int maxn = 4000; int A[maxn], B[maxn], C[maxn], D[maxn], CD[maxn * maxn]; int main() { int n; mr(n); for (int i = 0; i < n; ++i) mr(A[i]), mr(B[i]), mr(C[i]), mr(D[i]); for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j) CD[i * n + j] = C[i] + D[j]; sort(CD, CD + n * n); long long ans = 0; for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j) ans += upper_bound(CD, CD + n * n, -A[i] - B[j]) - lower_bound(CD, CD + n * n, -A[i] - B[j]); mw(ans, '\n'); return 0; } Followed by: Post your reply here: |
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