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题意：给出一个长度在 100 000 以内的正整数序列，大小不超过 10^ 12。求一个连续子序列，使得在所有的连续子序列中，
它们的GCD值乘以它们的长度最大。
析：暴力枚举右端点，然后在枚举左端点时，我们对gcd相同的只保留一个，那就是左端点最小的那个，只有这样才能保证是最大，然后删掉没用的。
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>
#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 = 1e5 + 5;
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -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); }
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;
}
LL a[maxn];
struct node{
int posi, posj;
LL val;
bool operator < (const node &p) const{
return val < p.val || (val == p.val && posi < p.posi);
}
node(int p, int q, LL x) : posi(p), val(x), posj(q) { }
};
vector<node> v;
vector<node> :: iterator it, it1;
int main(){
int T; cin >> T;
while(T--){
scanf("%d", &n);
for(int i = 0; i < n; ++i) scanf("%lld", a+i);
v.clear();
LL ans = 0;
for(int i = 0; i < n; ++i){
ans = Max(ans, a[i]);
for(int j = 0; j < v.size(); ++j){
ans = Max(ans, v[j].val * (v[j].posj-v[j].posi+1));
v[j].val = gcd(v[j].val, a[i]);
v[j].posj = i;
}
v.push_back(node(i, i, a[i]));
sort(v.begin(), v.end());
it = v.begin();
++it;
while(it != v.end()){
it1 = it; --it1;
if(it1->val == it->val) it = v.erase(it);
else ++it;
}
}
for(int i = 0; i < v.size(); ++i)
ans = Max(ans, v[i].val * (v[i].posj-v[i].posi+1));
printf("%lld\n", ans);
}
return 0;
}
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5898304.html
