POJ2689 Prime Distance

　　求相连素数最小距离和最大距离。

　　由于n很大，但是U-L较小，所以筛素数时只筛到sqrt(n)就可以了，利用这个已经筛出的素数表把[U,L]之间的素数筛出来。

View Code

 1 #include <stdio.h>
 2 #include <math.h>
 3 #include <stdlib.h>
 4 #include <string.h>
 5 #define LL long long
 6 #define N 50000
 7 #define M 1000001
 8 
 9 LL base[N], res[M];
10 LL l, u, k;
11 bool isprime[N], isres[M];
12 LL c1, c2, d1, d2;
13 
14 void init()  // 筛素数
15 {
16     LL i, j;
17     memset(isprime, false, sizeof isprime);
18     for(i = 2, k = 0; i < N; i ++)
19     {
20         if(!isprime[i])
21         {
22             base[k ++] = i;
23             for(j = i << 1; j < N; j += i)
24                 isprime[j] = true;
25         }
26     }
27 }
28 
29 void solve()
30 {
31     LL i, j, cnt;
32     memset(isres, false, sizeof isres);
33     for(i = 0; i < k; i ++)
34     {
35         LL tmp = l / base[i];      // 快速定位
36         while(tmp * base[i] < l || tmp <= 1)
37             ++ tmp;
38         for(j = tmp * base[i]; j <= u; j += base[i])
39             isres[j - l] = true;
40     }
41     if(l == 1)
42         isres[0] = true;
43     for(cnt = 0, i = 0; i <= u - l; i ++)
44     {
45         if(!isres[i])
46             res[cnt ++] = i + l;
47     }
48     if(cnt <= 1)
49         puts("There are no adjacent primes.");
50     else
51     {
52         LL max = 0, min = 0x7fffffff;
53         for(i = 1; i < cnt; i ++)
54         {
55             if(res[i] - res[i - 1] < min)
56             {
57                 min = res[i] - res[i - 1];
58                 c1 = res[i - 1];
59                 c2 = res[i];
60             }
61             if(res[i] - res[i - 1] > max)
62             {
63                 max = res[i] - res[i - 1];
64                 d1 = res[i - 1];
65                 d2 = res[i];
66             }
67         }
68         printf("%I64d,%I64d are closest, %I64d,%I64d are most distant.\n", c1, c2, d1, d2);
69     }
70 }
71 
72 int main()
73 {
74     init();
75     while(scanf("%lld%lld", &l, &u) != EOF)
76         solve();
77     return 0;
78 }