BZOJ3085: 反质数加强版SAPGAP

用质因子来表示数

先得到一个最小的$K$满足$2^K>p_i$

由$2^{q_1}{p_i}^{q_i}>2^{q_1+K-1}p_i^{q_i-1}$可以得到

$(q_1+1)cdot(q_i+1)>(q_1+K)cdot(q_i)$

整理可得$q_i<frac{q_1+1}{K-1}$

用类似的方法可以得到$q_1 < 2K-1$

搜索+上面两个剪枝即可通过本题。

#include <bits/stdc++.h>
using namespace std;
#define LL long long
struct bign{
    int a[30], m;
    bign(char *s) {
        m = 1;
        memset(a, 0, sizeof(a));
        int n = strlen(s);
        for(int i = 0; i < n; ++ i) {
            if(i) {
                if((n - i) % 4 == 0) ++ m;
            }
            a[m] = a[m] * 10 + s[i] - '0';
        }
        for(int i = 1; i <= m / 2; ++ i) {
            swap(a[i], a[m - i + 1]);
        }
    }
    bign(int x) {
        memset(a, 0, sizeof(a));
        a[1] = x; m = 1;
    }
    bign() {
        memset(a, 0, sizeof(a));
        m = 0;
    }
    inline bign operator * (const bign& rhs) const {
        bign ret = bign(0);
        ret.m = m + rhs.m - 1;
        for(int i = 1; i <= m; ++ i) {
            for(int j = 1; j <= rhs.m; ++ j) {
                ret.a[i + j - 1] += a[i] * rhs.a[j];
                ret.a[i + j] += ret.a[i + j - 1] / 10000;
                ret.a[i + j - 1] %= 10000;
            }
        }
        if(ret.a[m + rhs.m]) ++ ret.m;
        return ret;
    }
    inline bool operator < (const bign& rhs) const {
        if(m > rhs.m) return false;
        if(m < rhs.m) return true;
        for(int i = m; i >= 1; -- i) {
            if(a[i] < rhs.a[i]) return true;
            if(a[i] > rhs.a[i]) return false;
        }
        return false;
    }
    inline bool operator == (const bign& rhs) const {
        if(m > rhs.m) return false;
        if(m < rhs.m) return false;
        for(int i = m; i >= 1; -- i) {
            if(a[i] < rhs.a[i]) return false;
            if(a[i] > rhs.a[i]) return false;
        }
        return true;
    }
    inline bool operator <= (const bign& rhs) const {
        return (*this == rhs) || (*this < rhs);
    }
    inline void print() {
        printf("%d", a[m]);
        for(int i = m - 1; i >= 1; -- i) {
            printf("%04d", a[i]);
        }
    }
};
int prime[] =
{
    1,  2,  3,  5,  7,
    11, 13, 17, 19, 23,
    29, 31, 37, 41, 43,
    47, 53, 59, 61, 67,
    71, 73, 79, 83, 89,
    97, 101,103,107,109,
    113,127,131,137,139,
    149,151,157,163,167,
    173,179,181,191,193,
    197,199,211,223,227,
    229,233,239,241,251
};
int K[] =
{
    1,2,2,3,3,
    4,4,5,5,5,
    5,5,6,6,6,
    6,6,6,6,7,
    7,7,7,7,7,
    7,7,7,7,7,
    7,7,8,8,8,
    8,8,8,8,8,
    8,8,8,8,8,
    8,8,8,8,8,
    8,8,8,8,8
};
bign n, Ans;
LL mx;
char s[110];
int q1;
inline void dfs(int x, bign now, LL cnt, int lim) {
    if(cnt > mx || cnt == mx && now < Ans) {
        Ans = now; mx = cnt;
    }
    if(x != 1) {
        lim = min(lim, q1 / (K[x] - 1));
    }
    LL tmp = cnt;
    for(int i = 1; i <= lim; ++ i) {
        if(x == 1) q1 = i;
        tmp += cnt;
        now = now * bign(prime[x]);
        if(n < now) return;
        dfs(x + 1, now, tmp, i);
    }
}
int main() {
    scanf("%s", s);
    n = bign(s);
    if(n == bign(1)) {
        return puts("1"), 0;
    }
    int k = 0;
    bign tmp = bign(1);
    while(tmp <= n) {
        tmp = tmp * bign(prime[++ k]);
    }
    dfs(1, bign(1), 1,  2 * K[k] - 2);
    Ans.print();
}