洛谷 p2618 数字工程 记忆化搜索_ 线性筛

我们在线筛的同时处理出每个数的所有质因子，记忆化搜索的时候直接枚举质因子即可。

时间复杂度为 O(nlogn)O(nlogn)O(nlogn)

Code:

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;
const int maxn = 1000000 + 10;
vector<int>G[maxn];
int vis[maxn];
int prime[maxn], n, dp[maxn];
inline void init(){
    int cnt = 0;
    for(int i = 2;i <= n; ++i){
        if(!vis[i]) prime[++cnt] = i;
        for(int j = 1;j <= cnt && i * prime[j] <= n; ++j){
            vis[i * prime[j]] = prime[j];
            if(i % prime[j] == 0) break;
        }
        int k = i;
        while(vis[k]){
            G[i].push_back(vis[k]);
            int u = vis[k];
            while(k % u == 0) k /= u;
        }
        if(k != 1)G[i].push_back(k);
    }
}
int d(int u){
    if(dp[u] != -1) return dp[u];
    dp[u] = maxn;
    for(int i = 1;u - i >= 1 && i <= 6; ++i){
        dp[u] = min(dp[u], d(u - i) + i);
        if(!vis[u - i]) break;
    }
    int siz = G[u].size();
    for(int i = 0;i < siz; ++i)
        dp[u] = min(dp[u], d(u / G[u][i]) + 1);
    return dp[u];
}
int main(){
    n = maxn - 3;
    init();
    memset(dp, -1, sizeof(dp));
    for(int i = 2;i <= n; ++i) if(!vis[i]) dp[i] = 1;
    dp[1] = 0;
    int h;
    while(scanf("%d",&h) != EOF)
        printf("%d\n",d(h));
    return 0;
}