传送：UVA – 10487
10487 Closest Sums
Given is a set of integers and then a sequence of queries. A query gives you a number and asks to finda sum of two distinct numbers from the set, which is closest to the query number.
InputInput contains multiple cases.Each case starts with an integer n (1 < n ≤ 1000), which indicates, how many numbers are in theset of integer. Next n lines contain n numbers. Of course there is only one number in a single line. Thenext line contains a positive integer m giving the number of queries, 0 < m < 25. The next m linescontain an integer of the query, one per line.Input is terminated by a case whose n = 0. Surely, this case needs no processing.
OutputOutput should be organized as in the sample below. For each query output one line giving the queryvalue and the closest sum in the format as in the sample. Inputs will be such that no ties will occur.
Sample Input531217333431513031233123312334560
Sample OutputCase 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.
思路：首先先输入一堆数，然后排个序，再算出所有的不同的两个数的和，存入另一个数组（用set判重），对该数组排序，然后二分查找结果，注意这里数组要开大点，因为最大1000 *（1000 + 1）/ 2有500500，这里RE了一次
AC代码：
#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <set>#define LL long long#define INF 0x3fffffffusing namespace std;int n, m;int a[501000];int sum;int main() {int cas = 1;while(scanf("%d", &n), n) {int tmp[1005];for(int i = 0; i < n; i++) {scanf("%d", &tmp[i]);}sort(tmp, tmp + n);sum = 0;set<int> se;for(int i = 0; i < n – 1; i++) {for(int j = i + 1; j < n; j++) {int t = tmp[i] + tmp[j];if(!se.count(t)) {se.insert(t);a[sum++] = t;}}}sort(a, a + sum);//for(int i = 0; i < sum; i++) printf("%d\n", a[i]);printf("Case %d:\n", cas ++);scanf("%d", &m);while(m–) {int tt;scanf("%d", &tt);if(tt >= a[sum – 1]) {printf("Closest sum to %d is %d.\n", tt, a[sum – 1]);}else if(tt <= a[0]) {printf("Closest sum to %d is %d.\n", tt, a[0]);}else {int fir = 0, fin = sum – 1;while(fir < fin) {int m = fir + (fin – fir) / 2;if(a[m] < tt) fir = m + 1;else fin = m;}int ans;if(abs(a[fin] – tt) > abs(a[fin – 1] – tt)) ans = a[fin – 1];else ans = a[fin];printf("Closest sum to %d is %d.\n", tt, ans);}}}return 0;}
，放弃那些不愿放弃的，容忍那些不可容忍的。
