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。

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 #define LL long long
  4 struct bign{
  5     int a[30], m;
  6     bign(char *s) {
  7         m = 1;
  8         memset(a, 0, sizeof(a));
  9         int n = strlen(s);
 10         for(int i = 0; i < n; ++ i) {
 11             if(i) {
 12                 if((n - i) % 4 == 0) ++ m;
 13             }
 14             a[m] = a[m] * 10 + s[i] - '0';
 15         }
 16         for(int i = 1; i <= m / 2; ++ i) {
 17             swap(a[i], a[m - i + 1]);
 18         }
 19     }
 20     bign(int x) {
 21         memset(a, 0, sizeof(a));
 22         a[1] = x; m = 1;
 23     }
 24     bign() {
 25         memset(a, 0, sizeof(a));
 26         m = 0;
 27     }
 28     inline bign operator * (const bign& rhs) const {
 29         bign ret = bign(0);
 30         ret.m = m + rhs.m - 1;
 31         for(int i = 1; i <= m; ++ i) {
 32             for(int j = 1; j <= rhs.m; ++ j) {
 33                 ret.a[i + j - 1] += a[i] * rhs.a[j];
 34                 ret.a[i + j] += ret.a[i + j - 1] / 10000;
 35                 ret.a[i + j - 1] %= 10000;
 36             }
 37         }
 38         if(ret.a[m + rhs.m]) ++ ret.m;
 39         return ret;
 40     }
 41     inline bool operator < (const bign& rhs) const {
 42         if(m > rhs.m) return false;
 43         if(m < rhs.m) return true;
 44         for(int i = m; i >= 1; -- i) {
 45             if(a[i] < rhs.a[i]) return true;
 46             if(a[i] > rhs.a[i]) return false;
 47         }
 48         return false;
 49     }
 50     inline bool operator == (const bign& rhs) const {
 51         if(m > rhs.m) return false;
 52         if(m < rhs.m) return false;
 53         for(int i = m; i >= 1; -- i) {
 54             if(a[i] < rhs.a[i]) return false;
 55             if(a[i] > rhs.a[i]) return false;
 56         }
 57         return true;
 58     }
 59     inline bool operator <= (const bign& rhs) const {
 60         return (*this == rhs) || (*this < rhs);
 61     }
 62     inline void print() {
 63         printf("%d", a[m]);
 64         for(int i = m - 1; i >= 1; -- i) {
 65             printf("%04d", a[i]);
 66         }
 67     }
 68 };
 69 int prime[] =
 70 {
 71     1,  2,  3,  5,  7,
 72     11, 13, 17, 19, 23,
 73     29, 31, 37, 41, 43,
 74     47, 53, 59, 61, 67,
 75     71, 73, 79, 83, 89,
 76     97, 101,103,107,109,
 77     113,127,131,137,139,
 78     149,151,157,163,167,
 79     173,179,181,191,193,
 80     197,199,211,223,227,
 81     229,233,239,241,251
 82 };
 83 int K[] =
 84 {
 85     1,2,2,3,3,
 86     4,4,5,5,5,
 87     5,5,6,6,6,
 88     6,6,6,6,7,
 89     7,7,7,7,7,
 90     7,7,7,7,7,
 91     7,7,8,8,8,
 92     8,8,8,8,8,
 93     8,8,8,8,8,
 94     8,8,8,8,8,
 95     8,8,8,8,8
 96 };
 97 bign n, Ans;
 98 LL mx;
 99 char s[110];
100 int q1;
101 inline void dfs(int x, bign now, LL cnt, int lim) {
102     if(cnt > mx || cnt == mx && now < Ans) {
103         Ans = now; mx = cnt;
104     }
105     if(x != 1) {
106         lim = min(lim, q1 / (K[x] - 1));
107     }
108     LL tmp = cnt;
109     for(int i = 1; i <= lim; ++ i) {
110         if(x == 1) q1 = i;
111         tmp += cnt;
112         now = now * bign(prime[x]);
113         if(n < now) return;
114         dfs(x + 1, now, tmp, i);
115     }
116 }
117 int main() {
118     scanf("%s", s);
119     n = bign(s);
120     if(n == bign(1)) {
121         return puts("1"), 0;
122     }
123     int k = 0;
124     bign tmp = bign(1);
125     while(tmp <= n) {
126         tmp = tmp * bign(prime[++ k]);
127     }
128     dfs(1, bign(1), 1,  2 * K[k] - 2);
129     Ans.print();
130 }

 

posted @ 2018-08-05 21:07  iamunstoppable  阅读(214)  评论(0)  编辑  收藏  举报