UVa1152 4 Values whose Sum is 0 (Midway encounter method)

Links: http://vjudge.net/problem/36014

Analysis: First enumerate A and B, put all a+b records in an ordered array, and then enumerate C and D, check-c-d how many methods are written in the form of a+b (binary search).

1#include <cstdio>2#include <algorithm>3 using namespacestd;4 5 Const intMAXN =4000+5;6 7 intA[MAXN], B[MAXN], C[MAXN], D[MAXN];8 intAB[MAXN * MAXN], CD[MAXN *MAXN];9 Ten intMain () { One     intT; Ascanf"%d", &T); -      while(t--) { -         intN; thescanf"%d", &n); -          for(inti =0; I < n; i++) scanf ("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]); -          for(inti =0; I < n; i++) -              for(intj =0; J < N; J + +) Ab[i * n + j] = A[i] +B[j]; +          for(inti =0; I < n; i++) -              for(intj =0; J < N; J + +) Cd[i * n + j] =-(C[i] +d[j]); +Sort (CD, CD + n *n); A         Long LongAns =0; at          for(inti =0; I < n * N; i++) -Ans + = Upper_bound (CD, CD + N * N, Ab[i])-Lower_bound (CD, CD + n *N, Ab[i]); -printf"%lld\n", ans); -         if(T) printf ("\ n"); -     } -     return 0; in}

UVa1152 4 Values whose Sum is 0 (Midway encounter method)