B. Maximum Value (Codeforces Round #276 (div1 ),

B. Maximum Value (Codeforces Round #276 (div1 ),
B. Maximum Valuetime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

You are given a sequenceAConsistingNIntegers. Find the maximum possible value of (integer remainderAIDividedAJ), Where 1 limit ≤ limitI, Bytes,JLimit ≤ limitNAndAILimit ≥ limitAJ.

Input

The first line contains integerN-The length of the sequence (1 sequence ≤ sequenceNLimit ≤ limit 2-105 ).

The second line containsNSpace-separated integersAI(1 digit ≤ DigitAILimit ≤ limit 106 ).

Output

Print the answer to the problem.

Sample test (s) input

33 4 5

Output

2                                                                                                                             

Find the maximum value of a [j] % a [I] in a [j]

Because ai <1000000; can be hash, if a vertex does not exist, record the maximum value smaller than it.

For the value after finding the maximum modulo, take a [I] = I, that is, This vertex exists. The maximum value after the modulo operation must be I + 1 + k * I, in this way, each time I query is added,

Find the value closest to the maximum value.

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=2000000+100;int a[maxn];int main(){    int n;    scanf("%d",&n);    int x;    for(int i=0;i<n;i++)    {        scanf("%d",&x);        a[x]=x;    }    for(int i=0;i<maxn;i++)    {        if(a[i]!=i)        a[i]=a[i-1];    }    int ans=0;    for(int i=2;i<maxn;i++)    {        if(a[i]==i)        {            for(int j=i+i-1;j<maxn;j=j+i)            {                if(a[j]%i>ans&&a[j]>i)                ans=a[j]%i;            }        }    }    printf("%d\n",ans);    return 0;}

How can codeforces change the id color)

It's not Beijing time. It's four hours later. The color is not clear. I only know that 200 or 300 is blue.

ACM problems how does my code always time out ?? Http: // codeforcescom/contest/50/problem/B

Your algorithm needs optimization. In fact, you don't need to find the matching I j for each pair. You just need to find the number. Let me give you some simple code. You should understand it.

# Include <stdio. h>

Double table [256];
Char str [100001];

Void main ()
{
Char * p = str;
Int I;
Double n = 0;
Gets (str );
While (* p)
Table [* (p ++)] + = 1;
For (int I = 1; I <256; I ++)
N + = table [I] * table [I];
Printf ("%. 0lf", n );
}