POJ-2191 Mersenne Composite Numbers

题意：给定k，求所有满足小于2^k的所有梅森数（2^p（p为素数）-1被素因子分解的数）。

思路：此题一看就想到打表、先暴力打表，然后直接写即可。

View Code

 1 /**********************************暴力打表*****************************
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <string>
 7 #include <algorithm>
 8 #include <iostream>
 9 using namespace std;
10 
11 bool isprime(int n){
12     for(int i=2;i<n;i++)
13         if(n%i==0) return false;
14     return true;
15 }
16 
17 int main(){
18     
19 //    freopen("data.in","r",stdin);
20 //    freopen("data.out","w",stdout);
21     
22     for(int i=31;i<=61;i++){
23         if(isprime(i)){
24             long long k=1;
25             k<<=i;
26             k--;
27             for(long long j=2;j<=sqrt(k*1.0);j++){
28                 if(k%j==0){
29                     printf("%lld *",j);
30                     k/=j;
31                 }
32             }
33             if(k!=1) printf(" %lld",k);
34             k=1;
35             k<<=i;
36             k--;
37             printf(" = %lld = ( 2 ^ %d ) - 1\n",k,i);
38         }
39     }
40     return 0;
41 }
42 */ 
43 
44 
45 #include <cstdio>
46 #include <cmath>
47 #include <cstdlib>
48 #include <cstring>
49 #include <string>
50 #include <algorithm>
51 #include <iostream>
52 using namespace std;
53 
54 int a[10]={11,23,29,37,41,43,47,53,59}; 
55 char b[10][100]={
56     "23 * 89 = 2047 = ( 2 ^ 11 ) - 1",
57     "47 * 178481 = 8388607 = ( 2 ^ 23 ) - 1",
58     "233 * 1103 * 2089 = 536870911 = ( 2 ^ 29 ) - 1",
59     "223 * 616318177 = 137438953471 = ( 2 ^ 37 ) - 1",
60     "13367 * 164511353 = 2199023255551 = ( 2 ^ 41 ) - 1",
61     "431 * 9719 * 2099863 = 8796093022207 = ( 2 ^ 43 ) - 1",
62     "2351 * 4513 * 13264529 = 140737488355327 = ( 2 ^ 47 ) - 1",
63     "6361 * 69431 * 20394401 = 9007199254740991 = ( 2 ^ 53 ) - 1",
64     "179951 * 3203431780337 = 576460752303423487 = ( 2 ^ 59 ) - 1"
65 };
66 
67 int main(){
68     
69 //    freopen("data.in","r",stdin);
70 //    freopen("data.out","w",stdout);
71     
72     int n;
73     while(scanf("%d",&n)!=EOF){
74         for(int i=0;i<10;i++){
75             if(n>=a[i]) puts(b[i]);
76             else break;
77         }
78     }
79     return 0;
80 }

posted @ 2012-08-02 16:09 Hug_Sea 阅读(333) 评论(0) 收藏举报

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