lightoj Again Array Queries

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

巧妙暴力：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
const int MAXN=1e5+10;
int m[MAXN];
int cnt[1010];
void getans(int l,int r){
    if(r-l>=1000){//因为数字范围1-1000； 
        puts("0");return;
    }
    mem(cnt,0);
    for(int i=l;i<=r;i++)
        cnt[m[i]]++;
    int k=-1,ans=1000;
    for(int i=1;i<=1000;i++){
        if(cnt[i]>1){
            ans=0;break;
        }
        if(cnt[i]){
        if(k!=-1&&i-k<ans)ans=i-k;
        k=i;
        }
    }
    printf("%d\n",ans);
}
int main(){
    int T,n,q,flot=0;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&q);
        for(int i=0;i<n;i++)scanf("%d",m+i);
        printf("Case %d:\n",++flot);
        while(q--){
            int l,r;
            scanf("%d%d",&l,&r);
            getans(l,r);
        }
    }
    return 0;
}

 其实set写简单一些；

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<set>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
const int MAXN=1e5+10;
set<int>st;
int m[MAXN];
int cnt[1010];
void getans(int l,int r){
    if(r-l>=1000){//因为数字范围1-1000； 
        puts("0");return;
    }
    st.clear();
    mem(cnt,0);
    for(int i=l;i<=r;i++)
        st.insert(m[i]),cnt[m[i]]++;
    int ans=1000,k=-1;
    set<int>::iterator iter;
    for(iter=st.begin();iter!=st.end();iter++){
        if(cnt[*iter]>1){
            ans=0;break;
        }
        if(k!=-1&&*iter-k<ans)ans=*iter-k;
        k=*iter;
    }
    printf("%d\n",ans);
}
int main(){
    int T,n,q,flot=0;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&q);
        for(int i=0;i<n;i++)scanf("%d",m+i);
        printf("Case %d:\n",++flot);
        while(q--){
            int l,r;
            scanf("%d%d",&l,&r);
            getans(l,r);
        }
    }
    return 0;
}