二分查找(HDU 2141 )
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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): 27535 Accepted Submission(s): 6920
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 31 2 31 2 31 2 331410
Sample Output
Case 1:NOYESNO
Author
wangye
Source
HDU 2007-11 Programming Contest
Recommend
威士忌
思路:只是一道典型的二分查找题目;
二分。
输入有三个集合,要先合并两个为一,然后再对这个
合并出来的集合进行二分;否则不合并的话,直接用双层
for嵌套一个二分,会TLE的。
前者最坏情况:T*(25W+500*18*1000);
后者最坏情况:T*(1250W*10*1000),显然TLE。
#include"stdio.h"#include"string.h"#include"stdlib.h"int tot[3];int x[511];int y[511];int z[511];int d[251111],k;int cmp(const void *a,const void *b){ return *(int *)a-*(int *)b;}int main(){ int Case=1,n; int i,l; int aim,tmp,flag; int low,up,mid; while(scanf("%d%d%d",&tot[0],&tot[1],&tot[2])!=-1) { qsort(tot,3,sizeof(tot[0]),cmp); for(i=0;i<tot[0];i++) scanf("%d",&x[i]); for(i=0;i<tot[1];i++) scanf("%d",&y[i]); for(i=0;i<tot[2];i++) scanf("%d",&z[i]); qsort(x,tot[0],sizeof(x[0]),cmp); k=0; for(i=0;i<tot[1];i++) for(l=0;l<tot[2];l++) d[k++]=y[i]+z[l]; qsort(d,k,sizeof(d[0]),cmp); scanf("%d",&n); printf("Case %d:\n",Case++); while(n--) { scanf("%d",&aim); flag=0; for(i=0;i<tot[0];i++) { tmp=aim-x[i]; low=0;up=k-1;mid=(low+up)>>1; while(up-low>1) { if(d[mid]>tmp) up=mid; else if(d[mid]<tmp) low=mid; else {flag=1;break;} mid=(low+up)>>1; } if(d[low]==tmp||d[up]==tmp) flag=1; if(flag) break; } if(flag) printf("YES\n"); else printf("NO\n"); } } return 0;}
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