[51NOD1090] 3个数和为0（水题，二分）

题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1090

找到所有数的和，然后再原数组里二分找符合条件的第三个数。

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1010;
const int maxm = maxn * maxn;
typedef struct R {
  int x, y, z;
  R(){}
  R(int x, int y, int z) : x(x), y(y), z(z) {}
  bool operator == (R t) {
    return x == t.x && y == t.y && z == t.z;
  }
}R;
typedef struct S {
  int i, j;
  int s;
}S;

vector<R> ret;
int n, m;
int a[maxn];
S s[maxm];

bool cmp(R a, R b) {
  if(a.x == b.x) {
    if(a.y == b.y) return a.z < b.z;
    return a.y < b.y;
  }
  return a.x < b.x;
}

int main() {
  // freopen("in", "r", stdin);
  while(~scanf("%d", &n)) {
    m = 0; ret.clear();
    for(int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    sort(a, a+n);
    for(int i = 0; i < n; i++) {
      for(int j = i+1; j < n; j++) {
        s[m].i = i; s[m].j = j; s[m].s = a[i] + a[j];
        m++;
      }
    }
    for(int i = 0; i < m; i++) {
      int x = -s[i].s;
      int id = lower_bound(a, a+n, x) - a;
      if(a[id] == x && id != s[i].i && id != s[i].j) {
        int x = a[s[i].i], y = a[s[i].j], z = a[id];
        if(x > y) swap(x, y);
        if(x > z) swap(x, z);
        if(y > z) swap(y, z);
        ret.push_back(R(x,y,z));
      }
    }
    if(ret.size() == 0) {
      puts("No Solution");
      continue;
    }
    sort(ret.begin(), ret.end(), cmp);
    ret.erase(unique(ret.begin(), ret.end()), ret.end());
    for(int i = 0; i < ret.size(); i++) {
      printf("%d %d %d
", ret[i].x, ret[i].y, ret[i].z);
    }
  }
  return 0;
}