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。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int maxn = 1010; 5 const int maxm = maxn * maxn; 6 typedef struct R { 7 int x, y, z; 8 R(){} 9 R(int x, int y, int z) : x(x), y(y), z(z) {} 10 bool operator == (R t) { 11 return x == t.x && y == t.y && z == t.z; 12 } 13 }R; 14 typedef struct S { 15 int i, j; 16 int s; 17 }S; 18 19 vector<R> ret; 20 int n, m; 21 int a[maxn]; 22 S s[maxm]; 23 24 bool cmp(R a, R b) { 25 if(a.x == b.x) { 26 if(a.y == b.y) return a.z < b.z; 27 return a.y < b.y; 28 } 29 return a.x < b.x; 30 } 31 32 int main() { 33 // freopen("in", "r", stdin); 34 while(~scanf("%d", &n)) { 35 m = 0; ret.clear(); 36 for(int i = 0; i < n; i++) { 37 scanf("%d", &a[i]); 38 } 39 sort(a, a+n); 40 for(int i = 0; i < n; i++) { 41 for(int j = i+1; j < n; j++) { 42 s[m].i = i; s[m].j = j; s[m].s = a[i] + a[j]; 43 m++; 44 } 45 } 46 for(int i = 0; i < m; i++) { 47 int x = -s[i].s; 48 int id = lower_bound(a, a+n, x) - a; 49 if(a[id] == x && id != s[i].i && id != s[i].j) { 50 int x = a[s[i].i], y = a[s[i].j], z = a[id]; 51 if(x > y) swap(x, y); 52 if(x > z) swap(x, z); 53 if(y > z) swap(y, z); 54 ret.push_back(R(x,y,z)); 55 } 56 } 57 if(ret.size() == 0) { 58 puts("No Solution"); 59 continue; 60 } 61 sort(ret.begin(), ret.end(), cmp); 62 ret.erase(unique(ret.begin(), ret.end()), ret.end()); 63 for(int i = 0; i < ret.size(); i++) { 64 printf("%d %d %d ", ret[i].x, ret[i].y, ret[i].z); 65 } 66 } 67 return 0; 68 }