light oj 1100 - Again Array Queries

http://lightoj.com/volume_showproblem.php?problem=1100

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.

Sample Input

Output for Sample Input

2

5 3

10 2 3 12 7

0 2

0 4

2 4

2 1

1 2

0 1

Case 1:

1

1

4

Case 2:

1

题意：给你n个数和q次查询，每次查询区间[i, j]里面所有数 两两间的最小差值。

思路：由于所有数范围(1 —— 1000)我们可以用cnt记录区间里面每个数出现的次数，然后从前向后扫一遍所有数，只要某个数出现次数大于2，最小差值就为0；出现次数为1，那么逐个更新。

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>using namespace std;#define INF 0x3f3f3f3f#define N 300000int val[N], cnt[1001];int main(){    int t, nCase = 1;    scanf ("%d", &t);    while (t--)    {        int n, q, a, b;        scanf ("%d %d", &n, &q);        for (int i=0; i<n; i++)            scanf ("%d", &val[i]);        printf ("Case %d:\n", nCase++);        while (q--)        {            scanf ("%d %d", &a, &b);            if (b - a + 1 > 1000)                printf ("0\n");            else            {                int flag = 0, pre = 0, minx = INF;                memset (cnt, 0, sizeof (cnt));                for (int i=a; i<=b; i++)                    cnt[ val[i] ]++;                for (int i=1; i<=1000; i++)                {                    if (flag) break;                    if (cnt[i] >= 2)                    {                        flag = 1;                        break;                    }                    else if (cnt[i] == 1)                    {                        if (!pre) pre = i;                        else                        {                            minx = min (minx, i-pre);                            if (!minx)                            {                                flag = 1;                                break;                            }                            pre = i;                        }                    }                }            if (flag)                minx = 0;            printf ("%d\n", minx);            }        }    }    return 0;}