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1190 Reduced ID Numbers 求更高效的方法

2011-06-26 15:36 176 查看
1190.ReducedIDNumbers 
 
Total:1418Accepted:581Rating:3.5/5.0(4votes)
 
   
 TimeLimit:2sMemoryLimit:32MBDescriptionT.ChurteachesvariousgroupsofstudentsatuniversityU.EveryU-studenthasauniqueStudentIdentificationNumber(SIN).ASINsisanintegerintherange0≤s≤MaxSINwithMaxSIN=106-1.T.ChurfindsthisrangeofSINstoolargeforidentificationwithinhergroups.Foreachgroup,shewantstofindthesmallestpositiveintegerm,suchthatwithinthegroupallSINsreducedmodulomareunique.InputOnthefirstlineoftheinputisasinglepositiveintegerN,tellingthenumberoftestcases(groups)tofollow.EachcasestartswithonelinecontainingtheintegerG(1≤G≤300):thenumberofstudentsinthegroup.ThefollowingGlineseachcontainoneSIN.TheSINswithinagrouparedistinct,thoughnotnecessarilysorted.OutputForeachtestcase,outputonelinecontainingthesmallestmodulusm,suchthatallSINsreducedmodulomaredistinct.SampleInputCopysampleinputtoclipboard
2
1
124866
3
124866
111111
987651
SampleOutput
1
8
我的笨方法，暴力ac
#include<stdio.h>intjudge(intmod[300],intn);intmain(){intt,n,i,j,mod[300],mark=1;longintsin[300];scanf("%d",&t);while(t--){scanf("%d",&n);for(i=0;i<n;i++)scanf("%d",&sin[i]);if(n==1){printf("%d/n",1);continue;}for(i=2;mark==1;i++){for(j=0;j<n;j++)mod[j]=sin[j]%i;if(judge(mod,n))mark=0;}
printf("%d/n",i-1);mark=1;for(i=0;i<n;i++)mod[i]=sin[i]=0;}}intjudge(intmod[300],intn){intk,l;
for(k=0;k<n;k++)for(l=k;l<n-1;l++){if(mod[l+1]==mod[k])return0;}return1;}
太慢了，用了，0.36，真不知道前面的那些牛人怎样0.01s的，求方法啊
     
                                            

                                        

                        
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                         标签： 
                                                numbers
                                                integer
                                                each
                                                input
                                                output
                                                ini
                                                

                        

                    


                

            


            

                
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