题意：给定 n 个数，可以对所有的数进行缩小，问你找出和最大的数，使得这些数都能整除这些数中最小的那个数。
析：用前缀和来做，先统计前 i 个数中有有多少数，然后再进行暴力去找最大值，每次都遍历这一段就好。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e5 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){ return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){ return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn];
int sum[maxn];
int main(){
while(scanf("%d", &n) == 1){
int x;
memset(a, 0, sizeof a);
for(int i = 0; i < n; ++i){
scanf("%d", &x);
++a[x];
}
for(int i = 1; i <= 200000; ++i) sum[i] = sum[i-1] + a[i];
LL ans = 0;
for(int i = 1; i <= 200000; ++i) if(a[i]){
LL tmp = 0;
int j;
for(j = i-1; j <= 200000; j += i)
tmp += (LL)(sum[j] - sum[j-i]) * (j+1-i);
if(j > 200000) tmp += (LL)(sum[200000]-sum[j-i]) * (j+1-i);
ans = Max(ans, tmp);
}
cout << ans << endl;
}
return 0;
}