UVA, 543 Goldbach's Conjecture

题意：给你一个数，认为它能拆成两个素数之和的形式,能，则输出x = a + b ,不能则输出 "Goldbach‘s conjecture is wrong."

思路：打印素数表,x-prime[a]检查是否为素数

代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5
 6 #define MAXN 1000001
 7 #define ll
 8 int cnt;
 9 long long num;
10 bool isprime[MAXN];
11 long long prime[MAXN];
12
13 bool datecin()
14 {
15     if(scanf("%lld",&num)!=EOF)
16     {
17         if(num==0)
18             return false;
19         return true;
20     }
21     return false;
22 }
23
24 void getprime()
25 {
26     int i,j;
27     memset(isprime,1,sizeof(isprime));
28     isprime[0]=isprime[1]=0,cnt=0;
29     for(i=2;i*i<MAXN;i++)
30     {
31         if(isprime[i])
32         {
33             for(j=i*i;j<MAXN;j+=i)
34             {
35                 if(isprime[j])
36                 {
37                     isprime[j]=false;
38                 }
39             }
40
41         }
42     }
43     for(i=2;i<MAXN;i++)
44     {
45         if(isprime[i])
46             prime[cnt++]=i;
47     }
48 }
49
50 void showisp()
51 {
52     for(int i=0;i<MAXN;i++)
53     {
54         if(isprime[i])
55         cout<<i<<‘ ‘;
56     }
57     cout<<endl;
58     for(int i=0;i<cnt;i++)
59     {
60         cout<<prime[i]<<‘ ‘;
61     }
62 }
63
64 void datecal()
65 {
66     for(int i=0;i<cnt;i++)
67     {
68         if(num<prime[i])
69             break;
70         if(isprime[num-prime[i]])
71         {
72             cout<<num<<" = "<<prime[i]<<" + "<<num-prime[i]<<endl;
73             return;
74         }
75     }
76     cout<<"Goldbach‘s conjecture is wrong."<<endl;
77 }
78 int main()
79 {
80     getprime();
81     //showisp();
82     #ifdef ll
83     while(datecin())
84     {
85         datecal();
86     }
87     #endif // ll
88     return 0;
89 }

UVA, 543 Goldbach's Conjecture