Description

Killer Problem

You are given an array of N integers and
Q queries. Each query is a closed interval [l,
r]. You should find the minimum absolute difference between all pairs in that interval.

Input

First line contains an integer T (T10).
T sets follow. Each set begins with an integer
N (N200000). In the next line there are
N integers a_i (1a_i10⁴),
the number in the i-th cell of the array. Next line will contain
Q (Q10⁴).
Q lines follow, each containing two integers
l_i, r_i (1l_i,
r_iN,
l_i < r_i) describing the beginning and ending of of
i-th range. Total number of queries will be less than 15000.

Output

For the i-th query of each test output the minimum
| a_j–a_k| for
l_ij,
kr_i (jk)
a single line.

Sample Input

1
10
1 2 4 7 11 10 8 5 1 10000
4
1 10
1 2
3 5
8 10

Sample Output

0
1
3
4

题意：给定一个整数数组，对每组的询问L，R，输出区间差最小是多少

思路：排序后处理，但是这样会超时，因为题目说和不超过15000，所以最差也就只有15000个数

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
const int maxn = 200005;
const int inf = 0x3f3f3f3f;

int num[maxn];
int tmp[maxn];
int n, q;

void cal(int l, int r) {
	if (r - l + 1 > 15000) {
		printf("0\n");
		return;
	}
	int cnt = 0;
	for (int i = l; i <= r; i++)
		tmp[cnt++] = num[i];
	sort(tmp, tmp+cnt);
	int Min = inf;
	for (int i = 0; i < cnt-1; i++)
		Min = min(Min, tmp[i+1]-tmp[i]);
	printf("%d\n", Min);
}

int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		for (int i = 1; i <= n; i++)
			scanf("%d", &num[i]);
		scanf("%d", &q);
		int l, r;
		while (q--) {
			scanf("%d%d", &l, &r);
			cal(l, r);
		}
	}
	return 0;
}

UVA - 11898 Killer Problem