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.
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
这个题，直接O（n^3）肯定超时
仔细一想发现可以把时间降到O(n^2*logn)
但是我在实现的时候没实现好。还是超时了两次。
具体思路自然是先把a[i]+b[j]的所有情况存起来，存到sum数组里面
然后每次查找x-c[i]是否在sum中出现就OK
比较有意思的题目。
我的代码：
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
int a[505],b[505],c[505];
int sum[505*505];
int l,n,m,k;
bool judge(int x)
{
int left,right,mid;
left=1,right=k;
while(left<=right)
{
mid=(left+right)>>1;
if(sum[mid]>x)
right=mid-1;
else if(sum[mid]<x)
left=mid+1;
else
return true;
}
return false;
}
int main()
{
int i,j,q,x,cnt=1;
while(scanf("%d%d%d",&l,&n,&m)!=EOF)
{
for(i=1;i<=l;i++)
scanf("%d",&a[i]);
for(i=1;i<=n;i++)
scanf("%d",&b[i]);
for(i=1;i<=m;i++)
scanf("%d",&c[i]);
k=0;
for(i=1;i<=l;i++)
for(j=1;j<=n;j++)
{
k++;
sum[k]=a[i]+b[j];
}
sort(sum+1,sum+k+1);
scanf("%d",&q);
printf("Case %d:\n",cnt++);
while(q--)
{
scanf("%d",&x);
bool flag=true;
for(i=1;i<=m;i++)
{
if(judge(x-c[i]))
{
flag=false;
printf("YES\n");
break;
}
}
if(flag)
printf("NO\n");
}
}
return 0;
}