Gym 100299C && UVaLive 6582 Magical GCD (暴力+数论)

时间：2022-05-14 20:11:24

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。

UVaLive上的数据有问题，比赛时怎么也交不过，后来去别的oj交就过了。

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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