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UVa 1642 Magical GCD (暴力+数论)

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;}

 

 

UVa 1642 Magical GCD (暴力+数论)