BZOJ 1225: [HNOI2001] 求正整数( dfs + 高精度 )

15 < log₂50000 < 16, 所以不会选超过16个质数, 然后暴力去跑dfs, 高精度计算最后答案..

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#include<cstdio>

#include<algorithm>

#include<cstring>

#include<cmath>

 

using namespace std;

 

const int maxn = 50009;

const int n = 16;

const int p[n] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53};

 

int N, ans[n], g[n], ansn;

double cur = 1e30;

 

void dfs(int x, int lim, double res, int cnt) {

if(res > cur) return;

if(cnt == N) {

if(res < cur) {

ansn = x;

for(int i = 0; i < x; i++) ans[i] = g[i];

cur = res;

}

} else {

if(x == n) return;

for(int i = 0; i <= lim; i++) if(cnt * (i + 1) <= N) {

g[x] = i;

dfs(x + 1, i, res + i * log(p[x]), cnt * (i + 1));

}

}

}

 

struct Int {

static const int base = 10000;

static const int width = 4;

static const int maxn = 1000;

int s[maxn], n;

Int() {

n = 0;

memset(s, 0, sizeof s);

}

Int(int x) {

n = 0;

for(; x; x /= base) s[n++] = x % base;

}

Int operator = (const Int &o) {

n = o.n;

memcpy(s, o.s, sizeof(int) * n);

return *this;

}

Int operator * (const Int &o) const {

Int ret; ret.n = n + o.n - 1;

for(int i = 0; i < n; i++)

for(int j = 0; j < o.n; j++) 

ret.s[i + j] += s[i] * o.s[j];

for(int i = 0; i < ret.n; i++) if(ret.s[i] >= base) {

ret.s[i + 1] += ret.s[i] / base;

ret.s[i] %= base;

}

for(int &i = ret.n; ret.s[i]; i++) if(ret.s[i] >= base) {

ret.s[i + 1] += ret.s[i] / base;

ret.s[i] %= base;

}

return ret;

}

void write() {

int buf[width], c;

for(int i = n; i--; ) {

c = 0;

for(int t = s[i]; t; t /= 10) buf[c++] = t % 10;

if(i != n - 1)

for(int j = width - c; j--; ) putchar('0');

while(c--) putchar(buf[c] + '0');

}

puts("");

}

};

 

int main() {

scanf("%d", &N);

dfs(0, N - 1, 0, 1);

Int res = 1;

for(int i = 0; i < ansn; i++) {

Int _p = p[i];

for(int j = 0; j < ans[i]; j++)

res = res * _p;

}

res.write();

return 0;

}

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1225: [HNOI2001] 求正整数

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 576  Solved: 232
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Description

对于任意输入的正整数n，请编程求出具有n个不同因子的最小正整数m。例如：n=4，则m=6，因为6有4个不同整数因子1，2，3，6；而且是最小的有4个因子的整数。

Input

n（1≤n≤50000）

Output

m

Sample Input

4

Sample Output

6

HINT

Source

Dp