[BZOJ 3629][JLOI2014]聪明的燕姿-优快云博客

本文详细解析了一道编号为3629的算法题目，使用C++语言实现。通过深度优先搜索（DFS）算法寻找所有可能的质数解，并进行优化处理，确保了算法的高效性和准确性。

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爆搜题

/**************************************************************
    Problem: 3629
    User: MiEcoku
    Language: C++
    Result: Accepted
    Time:164 ms
    Memory:2468 kb
****************************************************************/
 
#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
 
const int maxn = 100010;
 
int t, cnt, n, ans[maxn], prim[maxn], book[maxn];
 
bool isprim(int x) {
    for ( register int i = 1; prim[i]*prim[i] <= x; ++ i)
        if( x % prim[i] == 0) return false;
    return true;
}
 
inline void dfs(int l, int op, int now) {
    if( now == 1) { ans[++ cnt]=  op; return;}
    if( now-1 >= prim[l] && isprim(now-1)) ans[++ cnt] = op * (now-1);
    for ( register int i = l; prim[i] * prim[i] <= now; ++ i) {
        for ( ll sum=1 + prim[i], p = prim[i]; sum <= now; p = p*prim[i], sum += p)
            if( !(now%sum)) dfs(i+1, op*p, now/sum);
    }
}
 
int main() {
    for ( register int i = 2; i <= 1e5; ++ i) {
        if( !book[i]) prim[++ t] = i;
        for ( register int k = 1; k <= t && prim[k]*i <= 1e5; ++ k) {
            book[i * prim[k]] = 1;
            if( i % prim[k] == 0) break;
        }
    }
    while ( ~scanf("%d", &n)) {
        cnt = 0; dfs(1, 1, n); sort(ans+1, ans+1+cnt);
        printf("%d\n", cnt);
        for ( register int i = 1; i <= cnt; ++ i) {
            printf(i == cnt ? "%d\n" : "%d ", ans[i]);
        }
    }
    return 0;
}