二分查找 UVa 10487 - Closest Sums，时间复杂度为O(2nlogn)

思路：

设给定的数为q，找出数组中的两个不相同的数（值可以相同）a、b，满足MIN{|q-(a+b)|}，当q-a=b时，得到最小值为0。

所以，可以先将数组排序，遍历a为数组中的数，二分查找数组中与q-a最为接近的数即为b。

排序和遍历的时间复杂度为均为O(nlogn)。

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

#define MAXN 1005
#define INF 0x7FFFFFFF

int a[MAXN];

int BSearch(int low, int high, int targ)
{
	if (targ <= a[low])
	{
		return low;
	}
	if (targ >= a[high])
	{
		return high;
	}
	while (low <= high)
	{
		int mid = (low + high) / 2;
		if (targ == a[mid])
		{
			return mid;
		}
		else if (targ < a[mid])
		{
			high = mid - 1;
		}
		else
		{
			low = mid + 1;
		}
	}
	return (targ - a[high]) < (a[low] - targ) ? high : low;
}

int ClosestSum(int n, int q)
{
	int closestSum, minDiff = INF;
	int i, j;
	for (i = 0; i < n-1; i++)
	{
		j = BSearch(i+1, n-1, q-a[i]);//查找范围[i+1,n-1]，去除重复结果，如i=1，j=2和i=2，j=1重复
		if (abs(a[i]+a[j]-q) < minDiff)
		{
			closestSum = a[i] + a[j];
			minDiff = abs(a[i]+a[j]-q);
		}
	}
	return closestSum;
}

int main(void)
{
	int n, m, caseID = 1;
	while (scanf("%d", &n) != EOF)
	{
		if (n == 0)
		{
			break;
		}

		int i;
		for (i = 0; i < n; i++)
		{
			scanf("%d", &a[i]);
		}

		sort(a, a+n);

		printf("Case %d:\n", caseID++);
		scanf("%d", &m);
		for (i = 0; i < m; i++)
		{
			int q;
			scanf("%d", &q);
			printf("Closest sum to %d is %d.\n", q, ClosestSum(n, q));
		}
	}
	return 0;
}