[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 = ;
 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 = ; ret.clear();
     for(int i = ; i < n; i++) {
       scanf("%d", &a[i]);
     }
     sort(a, a+n);
     for(int i = ; i < n; i++) {
       for(int j = i+; j < n; j++) {
         s[m].i = i; s[m].j = j; s[m].s = a[i] + a[j];
         m++;
       }
     }
     for(int i = ; 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() == ) {
       puts("No Solution");
       continue;
     }
     sort(ret.begin(), ret.end(), cmp);
     ret.erase(unique(ret.begin(), ret.end()), ret.end());
     for(int i = ; i < ret.size(); i++) {
       printf("%d %d %d\n", ret[i].x, ret[i].y, ret[i].z);
     }
   }
   return ;
 }