Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 3988 Accepted Submission(s): 997
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.
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".
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
在做此题的时候，如果单纯的用多重循环来实现会超时，正确的思路是先合并数组+排序+二分查找。下面是我的代码：
#include <stdio.h>
#include <stdlib.h>
int cmp(const void *a,const void *b)
{
return (*(int *)a-*(int *)b);
}
int main()
{
int f(int a[500*500],int x,int y,int z);
int i,j,m,n,s,t,x,y,z,k;
int a[500],b[500*500],c[500],d[1000];
z=1;
while(scanf("%d %d %d",&n,&m,&t)!=EOF)
{
y=0;
for(i=0;i<=n-1;i++)
{
scanf("%d",&a[i]);
}
for(i=0;i<=m-1;i++)
{
scanf("%d",&x);
for(j=0;j<=n-1;j++)
{
b[y]=a[j]+x;
y+=1;
}
}
for(i=0;i<=t-1;i++)
{
scanf("%d",&c[i]);
}
scanf("%d",&x);
for(i=0;i<=x-1;i++)
{
scanf("%d",&d[i]);
}
qsort(b,n*m,sizeof(b[0]),cmp);
printf("Case %d:\n",z);
for(i=0;i<=x-1;i++)
{
for(j=0;j<=t-1;j++)
{
s=d[i]-c[j];
k=f(b,0,n*m-1,s);
if(k==1)
{
break;
}
}
if(j==t)
{
printf("NO\n");
}else
{
printf("YES\n");
}
}
z+=1;
}
return 0;
}
int f(int a[500*500],int x,int y,int z)
{
int k,mid;
k=0;
while(x<=y)
{
mid=(x+y)/2;
if(a[mid]==z)
{
k=1;
break;
}else if(a[mid]>z)
{
y=mid-1;
}else
{
x=mid+1;
}
}
return (k);
}