学习二分查找+排序。一定用c++程序 sort用法;
http://acm.hdu.edu.cn/showproblem.php?pid=2141
Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others) Total Submission(s): 8557 Accepted Submission(s): 2230
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
3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10
Sample Output
Case 1:
NO
YES
NO
#include <iostream> #include <algorithm> using namespace std; int a[505],b[505],c[505],sab[250005]; int find( int a, int l , int h )//二分查找 { if( l > h ) return 0; int mid = ( l + h ) / 2; if ( sab[mid] == a ) return 1; else if( sab[mid] > a) find ( a, l, mid-1); else find ( a, mid+1, h); } int main() { int r,i,j,l,n,m,s,x,y,t=1; while(~scanf("%d%d%d",&l,&n,&m)) { r=0; for(i=0;i<l;i++) scanf("%d",&a[i]); for(i=0;i<n;i++) scanf("%d",&b[i]); for(i=0;i<m;i++) scanf("%d",&c[i]); for(i=0;i<l;i++) for(j=0;j<n;j++) sab[r++]=a[i]+b[j]; sort(sab,sab+r);//排序 scanf("%d",&s); printf("Case %d: ",t++); while(s--) { y=0; scanf("%d",&x); for(i=0;i<m;i++) { y=find(x-c[i],0,r-1); if(y==1) break; } if(y==1) printf("YES "); else printf("NO "); } } return 0; }