Wuyi nod 1421 Max MoD value

1421 Max MoD valuetitle Source: Codeforcesbase time limit: 1 seconds space limit: 131072 KB score: 80 Difficulty: 5-level algorithm problem

There is an an an array of n integers. Now you want to find two numbers (can be the same one)ai,aJ Makesai mod aJ Max and ai ≥ aJ.

Input

A single set of test data. The first line contains an integer n, which represents the size of the array A. (1≤n≤2*10^5) The second line has n spaces separated by an integer AI (1≤ai≤10^6).

Output

Output An integer representing the maximum mod value.

Input Example

33 4 5

Output Example

2

/*51 nod 1421 Max mod value problem: Select two numbers from the array to make A[I]%A[J] maximum solve: You can find A[i] a multiple of the closer to A[j] the greater. So enumerate the multiples of a[j] and then two points to find. At first I thought this would be t .... m+m/2+ ... M=o (MLGM) hhh-2016/09/16-16:33:10*/#pragma comment (linker, "/stack:124000000,124000000") #include <algorithm> #include <iostream> #include <cstdlib> #include <cstdio> #include <cstring> #include <vector > #include <math.h> #include <queue> #include <set> #include <map> #define Lson i<<1# Define Rson i<<1|1#define ll long Long#define CLR (A, B) memset (A,b,sizeof (a)) #define SCANFI (a) scanf ("%d", &a) # Define SCANFS (a) scanf ("%s", a) #define SCANFL (a) scanf ("%i64d", &a) #define SCANFD (a) scanf ("%lf", &a) #define Key _val ch[ch[root][1]][0] #define EPS 1e-7#define inf 0x3f3f3f3f3f3f3f3fusing namespace std;const ll mod = 1e9+7;const int ma    xn = 200010;const Double PI = ACOs ( -1.0); Template<class t> void Read (t&num) {char CH;    BOOL F=false; For (Ch=getchar (); ch< ' 0 ' | | Ch> ' 9 '; f= ch== '-', Ch=getchar ()); for (num=0; ch>= ' 0 ' &&ch<= ' 9 ';    num=num*10+ch-' 0 ', Ch=getchar ()); F && (num=-num);}        int stk[70], tp;template<class t> inline void print (T p) {if (!p) {puts ("0");    Return    } while (p) stk[++ TP] = p%10, p/=10;    while (TP) Putchar (stk[tp--] + ' 0 '); Putchar (' \ n ');}    int A[maxn];int Main () {int n;    Read (n);    for (int i =0; i < n; i++) {read (a[i]);    } sort (a,a+n);//int cnt = unique (a,a+n)-A;    int ans = 0;        for (int i =0; i < n; i++) {if (a[i] = = 1) continue;            for (int j = 1;j <= a[n-1]/a[i] + 1; j + +) {int pos = lower_bound (A+I,A+N,J * a[i])-A;            POS--;            if (pos > 0 && pos <= n && a[pos] > A[i]) ans = max (ans, a[pos]% a[i]); if (ans = = a[i]-1) break;//cout << J << "<< pos <<" "<<ans &L        t;< Endl; }} print (ANS); return 0;}

　　

Wuyi nod 1421 Max MoD value