1100 - Again Array Queries

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.

巧妙暴力：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cmath>
 6 using namespace std;
 7 #define mem(x,y) memset(x,y,sizeof(x))
 8 const int INF=0x3f3f3f3f;
 9 const double PI=acos(-1.0);
10 const int MAXN=1e5+10;
11 int m[MAXN];
12 int cnt[1010];
13 void getans(int l,int r){
14     if(r-l>=1000){//因为数字范围1-1000；
15         puts("0");return;
16     }
17     mem(cnt,0);
18     for(int i=l;i<=r;i++)
19         cnt[m[i]]++;
20     int k=-1,ans=1000;
21     for(int i=1;i<=1000;i++){
22         if(cnt[i]>1){
23             ans=0;break;
24         }
25         if(cnt[i]){
26         if(k!=-1&&i-k<ans)ans=i-k;
27         k=i;
28         }
29     }
30     printf("%d\n",ans);
31 }
32 int main(){
33     int T,n,q,flot=0;
34     scanf("%d",&T);
35     while(T--){
36         scanf("%d%d",&n,&q);
37         for(int i=0;i<n;i++)scanf("%d",m+i);
38         printf("Case %d:\n",++flot);
39         while(q--){
40             int l,r;
41             scanf("%d%d",&l,&r);
42             getans(l,r);
43         }
44     }
45     return 0;
46 }