NOI模拟题5 Problem A: 开场题

Solution

注意到(gcd)具有结合律:

[gcd(a, b, c) = gcd(a, gcd(b, c)) ]

因此我们从后往前, 对于每个位置(L), 找到每一段不同的(gcd(a_x, a_{x + 1}, cdots, a_R)). 我们注意到这样的(R)最多只有(log)段.

每个位置合并其后面一个位置的信息.

同时我们还维护序列的前缀异或和, 对于每个异或值, 都开一颗线段树来存储其出现的位置. 随便乱搞即可.

#include <cstdio>
#include <cctype>
#include <vector>
#include <algorithm>
#include <map>
 
typedef long long LL;
using namespace std;
namespace Zeonfai
{
    inline LL getLL()
    {
        LL a = 0, sgn = 1; char c;
        while (! isdigit(c = getchar())) if (c == '-') sgn *= -1;
        while (isdigit(c)) a = a * 10 + c - '0', c = getchar();
        return a * sgn;
    }
}
const int N = (int)1e5, INF = (int)2e9;
int n;
LL a[N + 1], sum[N + 1];
struct record
{
    int ed; LL val;
    inline record(int _ed, LL _val) { ed = _ed; val = _val; }
};
vector<record> bck[N + 1];
map<LL, int> mp;
struct segmentTree
{
    struct node
    {
        node *suc[2];
        inline node() { for (int i = 0; i < 2; ++ i) suc[i] = NULL; }
    }*rt;
    inline segmentTree() { rt = NULL; }
    node *insert(node *u, int L, int R, int pos)
    {
        if (u == NULL) u = new node;
        if (L == R) return u;
        if (pos <= L + R >> 1) u->suc[0] = insert(u->suc[0], L, L + R >> 1, pos);
        else u->suc[1] = insert(u->suc[1], (L + R >> 1) + 1, R, pos);
        return u;
    }
    inline void insert(int pos) { rt = insert(rt, 1, n, pos); }
    int find(node *u, int curL, int curR, int L, int R)
    {
        if (u == NULL) return INF;
        if (curL == curR) return curL;
        int res = INF, mid = curL + curR >> 1;
        if (L <= mid) res = min(res, find(u->suc[0], curL, mid, L, R));
        if (res == INF && R > mid) res = min(res, find(u->suc[1], mid + 1, curR, L, R));
        return res;
    }
    inline int find(int L, int R) { return find(rt, 1, n, L, R); }
}seg[N];
/* inline LL __gcd(LL a, LL b)
{
    if (a < b) swap(a, b);
    while (b)
    {
        LL tmp = b;
        b = a % b;
        a = tmp;
    }
    return a;
} */
inline vector<record>::iterator getPrevious(vector<record>::iterator p) { return -- p; }
int main()
{
 
#ifndef ONLINE_JUDGE
 
    freopen("start.in", "r", stdin);
    freopen("start.out", "w", stdout);
 
#endif
 
    using namespace Zeonfai;
    n = getLL();
    for (int i = 1; i <= n; ++ i) a[i] = getLL();
    for (int i = n; i; -- i)
    {
        bck[i].push_back(record(i, a[i])); int tot = 1;
        if (i != n)
        {
            for (vector<record>::iterator p = bck[i + 1].begin(); p != bck[i + 1].end(); ++ p)
                if (__gcd(p->val, a[i]) != bck[i][tot - 1].val)
                    bck[i].push_back(record(p->ed, __gcd(a[i], p->val))), ++ tot;
                else bck[i][tot - 1].ed = p->ed;
        }
    }
    sum[0] = 0; mp.clear(); int tot = 0;
    for (int i = 1; i <= n; ++ i)
    {
        sum[i] = sum[i - 1] ^ a[i];
        if (mp.find(sum[i]) == mp.end()) mp[sum[i]] = tot ++;
        seg[mp[sum[i]]].insert(i);
    }
    LL K = getLL();
    for (int i = 1; i <= n; ++ i)
    {
        for (vector<record>::iterator p = bck[i].begin(); p != bck[i].end(); ++ p)
        {
            if (K % p->val) continue;
            int res = INF;
            if (mp.find(K / p->val ^ sum[i - 1]) != mp.end())
                res = seg[mp[K / p->val ^ sum[i - 1]]].find(p == bck[i].begin() ? i : getPrevious(p)->ed + 1, p->ed);
            if (res != INF)
            {
                printf("%d %d
", i, res);
                return 0;
            }
        }
    }
    puts("no solution");
}