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  • CF A. DZY Loves Hash

    A. DZY Loves Hash
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    DZY has a hash table with p buckets, numbered from 0 to p - 1. He wants to insert n numbers, in the order they are given, into the hash table. For the i-th number xi, DZY will put it into the bucket numbered h(xi), where h(x) is the hash function. In this problem we will assume, that h(x) = x mod p. Operation a mod b denotes taking a remainder after division a by b.

    However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after the i-th insertion, you should output i. If no conflict happens, just output -1.

    Input

    The first line contains two integers, p and n (2 ≤ p, n ≤ 300). Then n lines follow. The i-th of them contains an integer xi (0 ≤ xi ≤ 109).

    Output

    Output a single integer — the answer to the problem.

    Sample test(s)
    Input
    10 5
    0
    21
    53
    41
    53
    
    Output
    4
    
    Input
    5 5
    0
    1
    2
    3
    4
    
    Output
    -1

    //题意就是找相等的数,输出第二个的位置。可是要是最先发现的。

    比如:10 5 1 2 2 2 1 输出是3而不是5,由于先找到2和2相等。假设仅仅用for循环,找到的是1和1相等输出是5. 第4个例子卡了非常久。没看懂题目。。。

    #include <iostream>
    using namespace std;
    int main()
    {   __int64 a[400];
        int n,t,i,j,p,k;
        while(scanf("%d%d",&p,&n)!=EOF)
        {   memset(a,0,sizeof(a));
            t=0;
            for(i=0;i<n;i++)
            {
                scanf("%I64d",&a[i]);
                a[i]=a[i]%p;
            }
            k=n;
            for(i=0;i<n-1;i++)
            {    
                for(j=i+1;j<n;j++)
                    if(a[i]==a[j])
                    {   
                        k=k<(j+1)?k:(j+1);
                            t=1;
                    }
                    
            }
            if(t==1)
                printf("%d
    ",k);
           if(t==0)
               printf("-1
    ");
        }
        return 0;
    }

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  • 原文地址:https://www.cnblogs.com/zhchoutai/p/8386360.html
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