随笔-141  评论-9  文章-3  trackbacks-0

这个代码好烂。。

/*
ID: lorelei3
TASK: crypt1
LANG: C++
*/


#include 
<fstream>
#include 
<stdlib.h>
#include 
<stdio.h>
#include 
<string.h>

using namespace std;

const int N = 10;

int s[N];
bool f[N];


int main(){
    
int a,b,c,d,e,i,n,ans=0;
    
int ai[3], bi[3], res[4];
    ifstream 
in("crypt1.in");
    ofstream 
out("crypt1.out");
    memset(f, 
falsesizeof(f));

    
in>>n;
    
for(i=0; i<n; ++i){
        
in>>s[i];
        f[s[i]] 
= true;
    }


    
    
for(a=0; a<n; ++a)
        
for(b=0; b<n; ++b)
            
for(c=0; c<n; ++c)
                
for(d=0; d<n; ++d){
                    
if(s[a]*s[d]>10)
                        
continue;
                    
for(e=0; e<n; ++e){
                        memset(ai, 
0sizeof(a));
                        memset(bi, 
0sizeof(b));
                        memset(res, 
0sizeof(res));

                        
int cc =0, bb=0;
                        
                        bb 
= (s[c]*s[e])%10;
                        cc 
= (s[c]*s[e])/10;

                        ai[
0= bb;

                        bb 
= (s[b]*s[e]+cc)%10;
                        cc 
= (s[b]*s[e]+cc)/10;

                        ai[
1= bb;
                        ai[
2= (s[a]*s[e]+cc);

                        
if(!(f[ai[0]] && f[ai[1]] && f[ai[2]]))
                            
continue;

                        
if(ai[2]>10)
                            
continue;

                        bb 
= (s[c]*s[d])%10;
                        cc 
= (s[c]*s[d])/10;

                        bi[
0= bb;

                        bb 
= (s[b]*s[d]+cc)%10;
                        cc 
= (s[b]*s[d]+cc)/10;

                        bi[
1= bb;
                        bi[
2= (s[a]*s[d]+cc);

                        
if(!(f[bi[0]] && f[bi[1]] && f[bi[2]]))
                            
continue;

                        
if(bi[2]>10)
                            
continue;
                        

                        res[
0]=ai[0];
                        
                        cc
=0;
                        bb
=(ai[1]+bi[0]+cc)%10;
                        cc
=(ai[1]+bi[0]+cc)/10;
                        res[
1]=bb;

                        bb
=(ai[2]+bi[1]+cc)%10;
                        cc
=(ai[2]+bi[1]+cc)/10;
                        res[
2]=bb;

                        res[
3]=bi[2]+cc;

                        
if(res[3]>10)
                            
continue;
                        
else{
                            
if(f[res[0]]&&f[res[1]]&&f[res[2]]&&f[res[3]]){
                                ans
++;
                            }

                        }

                    }

                }



    
out<<ans<<endl;
    
return 0;
}

posted on 2010-11-09 01:02 小阮 阅读(144) 评论(0)  编辑 收藏 引用 所属分类: USACO

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