codeforces 264B

题意很简单，给你n个严格上升的数字ai，
然后你要求出最大的子序列，满足相邻的数字不互质

我的思路比较奇葩，首先唯一分解每个ai，
然后通过他的质因子来寻找可能出现的转移的地方，然后dp[i] = max(dp[i], dp[k] + 1)
如果直接这样做会tle，然后我们想想，最靠前的边价值越小，不带转移的必要，所以我估算了下概率，舍掉80%的边直接不转移，然后就过了。。

////  Created by Matrix on 2016-01-22//  Copyright (c) 2015 Matrix. All rights reserved.//////#pragma comment(linker, "/STACK:102400000,102400000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <sstream>#include <set>#include <vector>#include <stack>#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x,begin())#define ll long long#define CLR(x) memset(x, 0, sizeof x)using namespace std;const int inf = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn = 1e5 + 10;const int maxv = 1e3 + 10;const double eps = 1e-9;int n;int a[maxn];vector <int> G[maxn];int dp[maxn];int main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt","w",stdout);#endif    while(scanf("%d", &n) != EOF) {        for(int i = 1; i <= 1e5; i++)            G[i].clear();        CLR(dp);        int ans = 0;        for(int i = 1; i <= n; i++) {            scanf("%d", a + i);            dp[i] = 1;            vector <int> factors;            int k = a[i];            int num = sqrt(a[i] + 0.5);            for(int j = 2; j <= num; j++) {                if(k % j == 0) {                    factors.push_back(j);                    while(k % j == 0) {                        k /= j;                    }                }                if(k == 1) break;            }            if(k != 1) factors.push_back(k);            for(k = 0; k < factors.size(); k++) {                num = factors[k];                int up = G[num].size();                up = up * 0.8;                for(int j = G[num].size() - 1; j >= up; j--) {                    dp[i] = max(dp[i], dp[G[num][j]] + 1);                }            }            for(k = 0; k < factors.size(); k++) {                num = factors[k];                G[num].push_back(i);            }            ans = max(ans, dp[i]);        }        cout << ans << endl;    }    return 0;}