Light oj--1100

Description

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

代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int num[1010];
int a[101000];
int main(){
	int t,cas;
	int n,q,u,v;
	cas=1;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&n,&q);
		for(int i=0;i<n;i++){
			scanf("%d",&a[i]);
		}
		printf("Case %d:\n",cas++);
		for(int i=1;i<=q;i++){
			scanf("%d%d",&u,&v);
			if(v-u>=1000){
				printf("0\n");
				continue;
			}
			int ans=0;
			int d=1005;//设一个初始差值 
			int flag=0;//判断是否为第一个数 
			memset(num,0,sizeof(num));
				for(int j=u;j<=v;j++){
					num[a[j]]++;//在区间[u,v]中每一个数出现的次数 
				}
				for(int k=1;k<=1000&&ans<=v-u;k++){
					if(num[k]>1){
					    d=0;
						break;	
					}
					else if(num[k]==1){
						ans++;
						if(flag==0){
							flag=k;//因为u!=v 
						}
						else{
							d=min(d,k-flag);
							flag=k;//不断变换两个数 
						}
						
					}
						
				}
				printf("%d\n",d);
			}
			
		}

	return 0;
}