UVA 10487 Closest Sums
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 integerm 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.
解题思路：现将每个数的两两之间的和存在一个数组里，，然后二分，要注意题目里输出中间是没有空行的（被坑了好久）。
#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>using namespace std;long long num[1005], ans[1005], sum[1000005];int main() {int Case = 1;int n, m;while (scanf("%d", &n), n) {for (int i = 0; i < n; i++) {scanf("%lld", &num[i]);}scanf("%d", &m);for (int i = 0; i < m; i++) {scanf("%lld", &ans[i]);}int cnt = 0;for (int i = 0; i < n; i++) {for (int j = i + 1; j < n; j++) {sum[cnt++] = num[i] + num[j];}}sort(sum, sum + cnt);printf("Case %d:\n", Case++);int Ans;for (int i = 0; i < m; i++) {int l = 0, r = cnt – 1;long long Ans = 0, mid = (l + r) / 2, min = 999999999;if (sum[mid] == ans[i]) {Ans = ans[i];}else if (sum[mid] < ans[i]) {while (mid <= r) {if (min > abs(ans[i] – sum[mid])) {min = abs(ans[i] – sum[mid]);Ans = sum[mid];}mid++;}}else {while (mid >= l) {if (min > abs(sum[mid] – ans[i])) {min = abs(sum[mid] – ans[i]);Ans = sum[mid];}mid–;}}printf("Closest sum to %lld is %lld.\n", ans[i], Ans);}}return 0;}
梦想，并不奢侈，只要勇敢地迈出第一步。
