【51nod】1061 最复杂的数 V2

题解

我是榜上最后一名= =

可能高精度用vector太慢了吧……什么破题= =

这道题很简单，如果高精度熟练代码……也很简单……然而，参数调了好久
我们发现质数的指数一定是，质数越小，指数越大，这个很显然我不说了

所以我们就用个优先队列BFS就好，队列按数从小到大排序，每次把队列的数取出来作为下一个我们需要的数（也就是大小递增且约数个数严格递增），删掉队列首比这个数约数个数小的数
然后用这个数再扩展一层质数，注意剪枝吧。。

预处理好后回答询问二分就行

质数大小开到85，搜出来的数的总量3810，在TLE的边缘试探……
vector写高精度是真的很慢……

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <ctime>
#include <vector>
#include <set>
//#define ivorysi
#define eps 1e-8
#define mo 974711
#define pb push_back
#define mp make_pair
#define pii pair<int,int>
#define fi first
#define se second
#define MAXN 100005
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
const int64 MOD = 1000000007;
const int BASE = 100000000,LEN = 8;
struct Bignum {
    vector<int> v;
    Bignum(int64 x = 0){
	*this = x;
    }
    Bignum operator = (int64 x) {
	v.clear();
	do {
	    v.pb(x % BASE);
	    x /= BASE;
	}while(x);
	return *this;
    }
    Bignum operator = (const string &str) {
	int x;
	v.clear();
	for(int i = str.length() ; i > 0 ; i -= LEN) {
	    int ed = i,st = max(i - LEN,0);
	    sscanf(str.substr(st,ed - st).c_str(),"%d",&x);
	    v.pb(x);
	}
	return *this;
    }
    friend Bignum operator * (const Bignum &a,const Bignum &b) {
	Bignum c;c.v.clear();
	for(int i = 1 ; i <= a.v.size() + b.v.size() ; ++i) c.v.pb(0);
	for(int i = 0 ; i < a.v.size() ; ++i) {
	    int g = 0;
	    for(int j = 0 ; j < b.v.size() ; ++j) {
		int64 x = 1LL * a.v[i] * b.v[j] + c.v[i + j] + g;
		c.v[i + j] = x % BASE;
		g = x / BASE;
	    }
	    int t = i + b.v.size();
	    while(g) {
		int64 x = c.v[t] + g;
		c.v[t] = x % BASE;
		g = x / BASE;
		++t;
	    }
	}
	for(int i = c.v.size() - 1 ; i > 0 ; --i) {
	    if(c.v[i] == 0) c.v.pop_back();
	    else break;
	}
	return c;
    }
    friend Bignum operator / (const Bignum &a,const int x) {
	Bignum c;c.v.clear();
	for(int i = 1 ; i <= a.v.size() ; ++i) c.v.pb(0);
	int g = 0;
	for(int i = a.v.size() - 1 ; i >= 0 ; --i) {
	    int64 y = 1LL * g * BASE + a.v[i];
	    c.v[i] = y / x;
	    g = y % x;
	}
	for(int i = c.v.size() - 1 ; i > 0 ; --i) {
	    if(c.v[i] == 0) c.v.pop_back();
	    else break;
	}
	return c;
    }
    friend bool operator < (const Bignum &a,const Bignum &b) {
	if(a.v.size() < b.v.size()) return true;
	else if(a.v.size() > b.v.size()) return false;
	else {
	    for(int i = a.v.size() - 1 ; i >= 0 ; --i) {
		if(a.v[i] < b.v[i]) return true;
		else if(a.v[i] > b.v[i]) return false;
	    }
	    return false;
	}
    }
    friend bool operator == (const Bignum &a,const Bignum &b) {
	if(a.v.size() != b.v.size()) return false;
	else {
	    for(int i = a.v.size() - 1 ; i >= 0 ; --i) {
		if(a.v[i] != b.v[i]) return false;
	    }
	    return true;
	}
    }
    friend bool operator > (const Bignum &a,const Bignum &b) {return b < a;}
    friend bool operator != (const Bignum &a,const Bignum &b) {return !(a == b);}
    friend bool operator <= (const Bignum &a,const Bignum &b) {return !(a > b);}
    friend bool operator >= (const Bignum &a,const Bignum &b) {return !(a < b);}
    void print() {
	int s = v.size() - 1;
	printf("%d",v[s]);
	--s;
	for(int i = s ; i >= 0 ; --i) {
	    printf("%08d",v[i]);
	}
    }
}N;
int T;

bool nonprime[100005];
int prime[100005],cnt;
const int P = 85;
struct node {
    Bignum num,val;
    int cnt[P + 5];
    node(Bignum _num = 0) {
	num = _num;
	memset(cnt,0,sizeof(cnt));
	val = 1;
    }
    friend node operator * (const node &a,int x) {
	node c;c.num = a.num * (Bignum)prime[x];
	memcpy(c.cnt,a.cnt,sizeof(c.cnt));
	c.cnt[x]++;
	c.val = a.val / (a.cnt[x] + 1) * (a.cnt[x] + 2);
	return c;
    }
    friend bool operator < (const node &a,const node &b) {
	return a.num < b.num;
    }
    friend bool operator == (const node &a,const node &b) {
	return a.num == b.num;
    }
}ans[4005];
int tot = 0;
set<node> S;
void Solve() {
    node p = node(1);
    int c = 3810;
    while(c--) {
	ans[++tot] = p;
	while(!S.empty()) {
	    node k = *S.begin();
	    if(k.val <= p.val) S.erase(S.begin());
	    else break;
	}
	S.insert(p * 1);
	node k = *S.begin();
	S.erase(S.begin());
	for(int i = 2 ; i <= cnt ; ++i) {
	    node t = p * i;
	    if(t.val <= k.val) continue;
	    S.insert(t);
	}
	p = k;	
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    for(int i = 2 ; i <= 10000 ; ++i) {
	if(!nonprime[i]) {
	    prime[++cnt] = i;
	    if(cnt >= P) break;
	    for(int j = 2 ; j <= 10000 / i ; ++j) {
		nonprime[i * j] = 1;
	    }
	}
    }
    ios::sync_with_stdio(false);
    Solve();
    cin>>T;
    string str;
    while(T--) {
	cin>>str;
	N = str;
	int L = 1,R = tot;
	while(L < R) {
	    int MID = (L + R + 1) >> 1;
	    if(ans[MID].num <= N) L = MID;
	    else R = MID - 1;
	}
	ans[L].num.print();
	putchar(' ');
	ans[L].val.print();
	putchar('
');
    }
    return 0;
}