Killer Problem （UVA 11898 ）

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 (T ≤ 10). T sets follow. Each set begins with an integer N (N ≤ 200000). In the next line there are N integers ai (1 ≤ ai ≤ 104 ), the number in the i-th cell of the array. Next line will contain Q (Q ≤ 104 ). Q lines follow, each containing two integers li , ri (1 ≤ li , ri ≤ N, li < ri) 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 |ajak| for li ≤ j, k ≤ ri (j ̸= k) 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

题解：因为给的N个数的范围很小，如果查询的区间的长度大于10000，那么区间一定有重复的数字，所以结果返回0，如果不是，把这个区间的所有出现的数记录在数组中，跑一遍[L,R]区间，求得相邻的出现的差值最小就是最后的答案。（根本不是线段树QTQ）

#include <bits/stdc++.h>

using namespace std;
const int maxn = 200005;
const int Max = 10004;
const int inf = 0x3f3f3f3f;

int a[maxn];
int b[Max];
int main()
{
    int t,n,m,i,l,r,ans,last;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i = 1; i <= n; i ++)scanf("%d",&a[i]);
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d%d",&l,&r);
            if(r - l + 1 >= 10000)   
            {
                printf("0
");
                continue;
            }
            else
            {
                memset(b,0,sizeof(b));
                for(i = l; i <= r; i ++)
                {
                    b[a[i]]++;
                    if(b[a[i]] > 1)
                    {
                        printf("0
");
                        break;
                    }
                }
                if(i <= r)continue;
                ans = inf;
                last = -inf;
                for(i = 1; i <= 10000; i ++)
                {
                    if(b[i]==1)
                    {
                        ans = min(ans,i - last);
                        last = i;
                    }
                }
                printf("%d
",ans);
            }
        }
    }
    return 0;
}