原题传送门: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): 10636 Accepted Submission(s): 2786
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
Author
wangye
Source
Recommend
分析:这道题目一共有三个序列集L N M 很显然直接暴力三个for肯定超时,就算两个for for 然后第三个用二分,这也是超时的。那么能用一个for和一个二分解决吗,显然是可以的。首先我们班L,N两个序列集合并为一个。定义一个新的 数组用来存储所有L,N集中和的情况,就是for for(A[n++] = L[i]+N[i]),然后再用遍历A集合,每一步在M集合中二分搜索是否有符合x-A[i]的值(利用这个公式就可以转换下了C = X-(A+B)),如果有就YES 木有就NO;
如果你这么天真地想,恭喜你你又得超时了。既然二分可以提高搜索速度,我们为什么不把A拿来二分搜索呢?A中的元素最多会有25W个啊!!如果二分就是10^3级别的。。搜索次数,大大减小时间。如果拿M来二分,你只能从500到20左右。
OJ也证明了这一点,代码如下:
//搜索 HDU2141 #include<cstdio> #include<algorithm> using namespace std; #define MAXN 510 int AL[MAXN]; int AN[MAXN]; int A[MAXN*MAXN]; int AM[MAXN]; int L,M,N,x,s,k,temp,n; bool res; bool binSearch(int target,int len){ int low = 0; int high = len - 1; int mid; while(low<high) { mid = (low+high)/2; if(A[mid] == target)return true; else if(target>A[mid]) low = mid+1; else high = mid; } return false; } int main() { k = 1; while(scanf("%d%d%d",&L,&N,&M)==3) { n = 0; res = false; for(int i=0;i<L;i++) scanf("%d",AL+i); for(int i=0;i<N;i++) scanf("%d",AN+i); for(int i=0;i<M;i++) scanf("%d",AM+i); for(int i=0;i<L;i++){ for(int j=0;j<N;j++){ A[n++]= AL[i]+AN[j]; } } sort(A,A+n); sort(AM,AM+M); scanf("%d",&s); printf("Case %d:\n",k++); while(s--){ scanf("%d",&x); for(int i=0;i<M;i++){ res = binSearch(x-AM[i],n); if(res){ printf("YES\n"); break; } } if(!res){ printf("NO\n"); } } } return 0; }