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参看资料：
https://blog.csdn.net/helloiamclh/article/details/49431025?utm_source=blogxgwz1
题目：
Given an array with n integers, and you are given two indices i and j (i ≠ j) in the array. You have to find two integers in the range whose difference is minimum. You have to print this value. The array is indexed from 0 to n-1.
Input
Input starts with an integer T (≤ 5), denoting the number of test cases.
Each case contains two integers n (2 ≤ n ≤ 105) and q (1 ≤ q ≤ 10000). The next line contains n space separated integers which form the array. These integers range in [1, 1000].
Each of the next q lines contains two integers i and j (0 ≤ i < j < n).
Output
For each test case, print the case number in a line. Then for each query, print the desired result.
Sample Input
2
5 3
10 2 3 12 7
0 2
0 4
2 4
2 1
1 2
0 1
Sample Output
Case 1:
1
1
4
Case 2:
1
题目大意：
n个数据，q个询问。每个询问给定 l ，r ，求区间 [ l , r ] 之内 任意两个数 的 差的绝对值 的 最小值。
解题思路：
暴力暴力暴力，用set装区间内的元素，以为不考虑重复元素的差值，错的离谱，。
用到桶的思想，桶内有两个元素，则最小差值为 0 ；否则，不断找寻最小距离的两个数。
实现代码：
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define MAXN 100010
#define MAX(a,b) a>b?a:b
#define MIN(a,b)a>b?b:a
#define INF 0xfffffff
using namespace std;
int a[MAXN],v[MAXN];
int fab(int a){return a>0?a:-a;
}int main(){int t,i,n,k,j,l,r;int cas=0;scanf("%d",&t);while(t--){scanf("%d%d",&n,&k);for(i=0;i<n;i++)scanf("%d",&a[i]);printf("Case %d:\n",++cas);while(k--){int ans=INF;scanf("%d%d",&l,&r);if(r-l+1>1000){printf("0\n");continue;}memset(v,0,sizeof(v));for(i=l;i<r;i++){if(v[a[i]]){ans=0;break;}v[a[i]]=1;for(j=i+1;j<=r;j++)ans=MIN(ans,fab(a[i]-a[j]));}printf("%d\n",ans);}}return 0;
}
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