UVA10487(二分）

Given is a set of integers and then a sequence of queries. A query gives you a number and asks to find a sum of two distinct numbers from the set, which is closest to the query number.
Input
Input contains multiple cases. Each case starts with an integer n (1 < n ≤ 1000), which indicates, how many numbers are in the set of integer. Next n lines contain n numbers. Of course there is only one number in a single line. The next line contains a positive integer m giving the number of queries, 0 < m < 25. The next m lines contain an integer of the query, one per line. Input is terminated by a case whose n = 0. Surely, this case needs no processing.
Output
Output should be organized as in the sample below. For each query output one line giving the query value and the closest sum in the format as in the sample. Inputs will be such that no ties will occur.
Sample Input
5 3 12 17 33 34 3 1 51 30 3 1 2 3 3 1 2 3 3 1 2 3 3 4 5 6 0
Sample Output
Case 1: Closest sum to 1 is 15. Closest sum to 51 is 51. Closest sum to 30 is 29. Case 2: Closest sum to 1 is 3. Closest sum to 2 is 3. Closest sum to 3 is 3. Case 3: Closest sum to 4 is 4. Closest sum to 5 is 5. Closest sum to 6 is 5.

题意：给一个数和一组数列，在数列选两个数组合出距离这个数最近的值。

收获:了解了unique函数。

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
using namespace std;

const int INF=0x3f3f3f3f;
const double eps=1e-10;
const double PI=acos(-1.0);
#define maxn 500
vector<int> a;
vector<int> a1;
vector<int> a2;
int n;
int main()
{
    int cas = 1;
    while(~scanf("%d", &n)&&n)
    {
        a.clear();
        a1.clear();
        int b;
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &b);
            a.push_back(b);
        }
        for(int i = 0; i < a.size(); i++)
            for(int j = i + 1; j < a.size(); j++)
            a1.push_back(a[i]+a[j]);
        sort(a1.begin(), a1.end());
//        a2.clear();
//        a2.push_back(a1[0]);
//        for(int i = 1; i < a1.size(); i++)  if(a1[i] != a1[i-1]) a2.push_back(a1[i]);
        vector<int>::iterator iter = unique(a1.begin(), a1.end());
        a1.erase(iter, a1.end());
        int m;
        printf("Case %d:\n", cas++);
        scanf("%d", &m);
        int c;
        for(int i = 0; i < m; i++)
        {
            scanf("%d", &c);
            vector<int>::iterator it;
            it = lower_bound(a1.begin(), a1.end(), c);
            if(*it == c) printf("Closest sum to %d is %d.\n", c, c);
            else
            {
                int ans = *it;
                if(it != a1.begin() && abs(*(it-1) - c) < abs(*it - c))
                ans = *(it - 1);
                printf("Closest sum to %d is %d.\n", c, ans);

            }
        }
    }
    return 0;
}