Can you find it?+合并字串+二分查找
2013-04-18 20:07
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Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)Total Submission(s): 7114 Accepted Submission(s): 1851
[align=left]Problem Description[/align]
Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.
[align=left]Input[/align]
There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent
the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.
[align=left]Output[/align]
For each case, firstly you have to print the case number as the form "Case d:", then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print "YES", otherwise print "NO".
[align=left]Sample Input[/align]
3 3 3 1 2 3 1 2 3 1 2 3 3 1 4 10
[align=left]Sample Output[/align]
Case 1: NO YES NO
[align=left]Author[/align]
wangye
[align=left]Source[/align]
HDU 2007-11 Programming Contest
[align=left]Recommend[/align]
威士忌
/* 思路:二分查找 技巧:合并其中两个串,再进行二分,不然超时 */ #include<iostream> using namespace std; int d[250010]; int cmp(const void *a,const void *b){ return *(int *)a-*(int *)b; } int main() { int l,n,m,count=1; while(cin>>l>>n>>m){ int a[502],b[502],c[502]; int i,j; for(i=0;i<l;i++) cin>>a[i]; for(i=0;i<n;i++) cin>>b[i]; for(i=0;i<m;i++) cin>>c[i]; int k=0; for(i=0;i<n;i++) for(j=0;j<m;j++) d[k++]=b[i]+c[j]; qsort(d,k,sizeof(d[0]),cmp); int s; cin>>s; cout<<"Case "<<count++<<":"<<endl; while(s--){ int x,flag=0; cin>>x; for(i=0;i<l;i++){ int num=x-(a[i]); int low=0,high=k,mid; while(low<=high){ mid=(low+high)/2; if(d[mid]<num) low=mid+1; else { if(d[mid]==num) break; else high=mid-1; } } if(num==d[mid]){ flag=1; cout<<"YES"<<endl; break; } } if(!flag) cout<<"NO"<<endl; } } return 0; }
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