Can you find it?-CFANZ编程社区

Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 17505 Accepted Submission(s): 4426

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

Case 1:
NO
YES
NO

C语言程序代码

给三个数L，N，M，接着给三组数，第一组是L个数，第二组是N个数

   
   		第三组是M个数，从这三组中各拿一个数相加。然后给一个数T，表示

   
   		接下来要输入T个数，判断它是否等于拿出的三个数的和。如是，输出

   
   		YES，否则输出NO。
解题思路::

   
   		用二分法做，先将前两组数两两相加把所有情况存放到一个数组中，对其进行排序

   
   		然后与要比较的数取差，最后与第三组比较，如相等则跳出，否则继续比较

   
   		（用二分法节省时间） 

   
   		 


*/
#include<stdio.h>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;
int a[510],b[510],c[510],d[250010];
int main(){
    int L,N,M,i,j,k,n,m,l,t,s,x,y,flag;
    s=0;
    while(~scanf("%d %d %d",&L,&N,&M))
    {
        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]);
            sort(c,c+M);

   
   		m=0;
        for(j=0;j<L;j++)
            for(k=0;k<N;k++)

   
   			d[m++]=a[j]+b[k];

   
   		sort(d,d+m);           
        scanf("%d",&t);    
        printf("Case %d:\n",++s);    
        for(i=0;i<t;i++)
        {
            scanf("%d",&x);
            flag=0;
            for(k=0;k<M;k++)
            {
                y=x-c[k];
                int le,r,mid;
                le=0;r=m-1;
                while(le<=r)
                {
                    mid=(le+r)/2;
                    if(d[mid]==y)
                    {
                        flag=1;
                        break;
                    }
                    if(d[mid]<y)
                        le=mid+1;
                    else
                        r=mid-1;
                }
                if(flag==1)
                    break;
            }
            if(flag)
                printf("YES\n");
            else
                printf("NO\n");                
        }

   
   	}
    return 0;
}