hdu杭电 2141 Can you find it? 【二分 N*logN】
2015-07-30 16:44
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Problem Description
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.
Input
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.
Output
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".
Sample Input
Sample Output
//ac代码
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.
Input
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.
Output
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".
Sample Input
3 3 3 1 2 3 1 2 3 1 2 3 3 1 4 10
Sample Output
Case 1: NO YES NO
//ac代码
//Can you find it #include<cstdio> #include<algorithm> using namespace std; int s[550*550]; int main() { int m,n,l;//表示abc int t,x; int i,k,j; int mid; int a[550],b[550],c[550]; int num=0; while(~scanf("%d%d%d",&m,&n,&l)) { for(i=0;i<m;++i) { scanf("%d",a+i); } for(i=0;i<n;++i) { scanf("%d",b+i); } for(i=0;i<l;++i) { scanf("%d",c+i); } int p=0; for(j=0;j<n;++j)//把b集合和c集合的数存到s数组中 { for(k=0;k<l;++k) { s[p++]=b[j]+c[k]; } } sort(s,s+p);//排序 printf("Case %d:\n",++num); scanf("%d",&t); while(t--) { int flag=0; scanf("%d",&x); for(i=0;i<m;++i) { int l,r; if(s[0]+a[i]==x||s[p-1]+a[i]==x) { flag=1; break; } /*else if(s[0]+a[i]>x || s[p-1]+a[i]<x) { continue; }*/ l=0;r=p-1; while(r>=l) { mid=(l+r)/2; if(s[mid]+a[i]==x) { flag=1; break; } else if(s[mid]+a[i]<x) l=mid+1; else if(s[mid]+a[i]>x) r=mid-1; } if(flag) break; } if(flag) printf("YES\n"); else printf("NO\n"); } } return 0; }
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