POJ 2549

Sumsets
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 8235   Accepted: 2260

Description

Given S, a set of integers, find the largest d such that a + b + c = d where a, b, c, and d are distinct elements of S.

Input

Several S, each consisting of a line containing an integer 1 <= n <= 1000 indicating the number of elements in S, followed by the elements of S, one per line. Each element of S is a distinct integer between -536870912 and +536870911 inclusive. The last line of input contains 0.

Output

For each S, a single line containing d, or a single line containing "no solution".

Sample Input

5
2 
3 
5 
7 
12
5
2 
16 
64 
256 
1024
0

Sample Output

12
no solution

Source

Waterloo local 2001.06.02
 
分成a + b  和 d - c两个集合
x = d - c    保存a + b 的所有值，二分查找x并判断合理性
 
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 
 6 using namespace std;
 7 
 8 struct node { int v,a,b; };
 9 const int MAX = 1005;
10 int N,len = 0;
11 int ele[MAX];
12 node y[MAX * MAX];
13 
14 bool cmp(node a,node b) { return a.v < b.v; }
15 
16 bool solve() {
17         for(int i = N - 1; i >= 0; --i) {
18                 for(int j = N - 1; j >= 0; --j) {
19                         if(i == j) continue;
20                         int d = ele[i] - ele[j];
21                        // printf("ele = %d d = %d\n",ele[i],d);
22                         int l = 0,r = len - 1;
23                         while(l < r) {
24                                 int mid = (l + r + 1) >> 1;
25                                 if(y[mid].v > d) r = mid - 1;
26                                 else l = mid;
27                         }
28                         if(y[l].v == d) {
29                                 for(int k = l; k < len; ++k) {
30                                         if(y[k].v == d && y[k].a != ele[j] && y[k].b != ele[j]
31                                            && y[k].a != ele[i] && y[k].b != ele[i]) {
32                                                 printf("%d\n",ele[i]);
33                                                 return true;
34                                         }
35                                 }
36                                 for(int k = l; k >= 0; --k) {
37                                         if(y[k].v == d && y[k].a != ele[j] && y[k].b != ele[j]
38                                            && y[k].a != ele[i] && y[k].b != ele[i]) {
39                                                 printf("%d\n",ele[i]);
40                                                 return true;
41                                         }
42 
43                                 }
44 
45                         }
46 
47                 }
48         }
49 
50         return false;
51 
52 }
53 
54 int main()
55 {
56    // freopen("sw.in","r",stdin);
57 
58     while(~scanf("%d",&N) && N ) {
59             for(int i = 0; i < N; ++i) scanf("%d",&ele[i]);
60 
61             sort(ele,ele + N);
62             N = unique(ele,ele + N) - ele;
63 
64 
65             len = 0;
66             for(int i = 0; i < N; ++i) {
67                     for(int j = i + 1; j < N; ++j) {
68                             y[len].v = ele[i] + ele[j];
69                             y[len].a = ele[i];
70                             y[len++].b = ele[j];
71                     }
72             }
73 
74             sort(y,y + len,cmp);
75 
76             if(!solve()) printf("no solution\n");
77     }
78     return 0;
79 }
View Code

 

posted @ 2014-04-02 21:57  hyx1  阅读(276)  评论(0)  编辑  收藏  举报