lightoj 1100 - Again Array Queries 【思维题】

1100 - Again Array Queries

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Time Limit: 3 second(s)Memory Limit: 32 MB

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

Notes

Dataset is huge, use faster I/O methods.

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SPECIAL THANKS: JANE ALAM JAN (DESCRIPTION, SOLUTION, DATASET)

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。

AC代码：

#include <cstdio>#include <cstring>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 100100using namespace std;int cnt[1001];int num[MAXN];int kcase = 1;void solve(){    int n, q;    scanf("%d%d", &n, &q);    for(int i = 0; i < n; i++)        scanf("%d", &num[i]);    printf("Case %d:\n", kcase++);    while(q--)    {        int a, b;        scanf("%d%d", &a, &b);        if(b - a + 1 > 1000)            printf("0\n");        else        {            memset(cnt, 0, sizeof(cnt));            for(int i = a; i <= b; i++)                cnt[num[i]]++;            int pre = 0;            int ans = INF;            bool flag = false;            for(int i = 1; i <= 1000; i++)            {                if(flag)                    break;                if(cnt[i] >= 2)                {                    flag = true;                    break;                }                else if(cnt[i] == 1)                {                    if(pre == 0)                        pre = i;                    else                    {                        ans = min(ans, i - pre);                        if(ans == 0)                            flag = true;                        pre = i;                    }                }            }            if(flag)                ans = 0;            printf("%d\n", ans);        }    }}int main(){    int t;    scanf("%d", &t);    while(t--){        solve();    }    return 0;}