Can you find it?
Time Limit : 10000/3000ms (Java/Other) Memory Limit : 32768/10000K (Java/Other)Total Submission(s) : 274 Accepted Submission(s) : 87
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
纳闷儿的是我注释掉的明明可以减轻程序运行的负担～但实际上确实Wrong answer~注释掉就ac了~~~
AC-code:
Case 1:
NO
YES
NO
#include<stdio.h>
#include<algorithm>
using namespace std;
int A[505],B[505],C[505],num[500*500*500+10];
int main()
{
int l,m,n,i,j,le,k,r,x,s,mid,p=1;
while(~scanf("%d%d%d",&l,&m,&n))
{
for(i=0;i<l;i++)
scanf("%d",&A[i]);
for(i=0;i<m;i++)
scanf("%d",&B[i]);
for(i=0;i<n;i++)
scanf("%d",&C[i]);
printf("Case %d:\n",p++);
scanf("%d",&s);
int len=0;
for(i=0;i<l;i++)
for(j=0;j<m;j++)
num[len++]=A[i]+B[j];
sort(num,num+len);
sort(C,C+n);
while(s--)
{
k=0;
scanf("%d",&x);
//if(x<num[0]+C[0]||x>num[len-1]+C[n-1])
//{
//printf("NO\n");
//continue;
//}
for(i=0;i<n;i++)
{
le=0;r=len-1;
while(le<=r)
{
mid=(le+r)/2;
if(x-num[mid]==C[i])
{
k=1;
break;
}
else if(x-num[mid]<C[i])
r=mid-1;
else
le=mid+1;
}
if(k)
break;
}
if(k)
printf("YES\n");
else
printf("NO\n");
}
}
return 0;
}