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; }
时间: 2024-10-10 05:16:52