lightoj Again Array Queries

1100 - Again Array Queries

PDF (English)

Statistics

Forum

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+;
 int m[MAXN];
 int cnt[];
 void getans(int l,int r){
     if(r-l>=){//因为数字范围1-1000；
         puts("");return;
     }
     mem(cnt,);
     for(int i=l;i<=r;i++)
         cnt[m[i]]++;
     int k=-,ans=;
     for(int i=;i<=;i++){
         if(cnt[i]>){
             ans=;break;
         }
         if(cnt[i]){
         if(k!=-&&i-k<ans)ans=i-k;
         k=i;
         }
     }
     printf("%d\n",ans);
 }
 int main(){
     int T,n,q,flot=;
     scanf("%d",&T);
     while(T--){
         scanf("%d%d",&n,&q);
         for(int i=;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 ;
 }

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+;
set<int>st;
int m[MAXN];
int cnt[];
void getans(int l,int r){
    if(r-l>=){//因为数字范围1-1000；
        puts("");return;
    }
    st.clear();
    mem(cnt,);
    for(int i=l;i<=r;i++)
        st.insert(m[i]),cnt[m[i]]++;
    int ans=,k=-;
    set<int>::iterator iter;
    for(iter=st.begin();iter!=st.end();iter++){
        if(cnt[*iter]>){
            ans=;break;
        }
        if(k!=-&&*iter-k<ans)ans=*iter-k;
        k=*iter;
    }
    printf("%d\n",ans);
}
int main(){
    int T,n,q,flot=;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&q);
        for(int i=;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 ;
}