Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 23679 Accepted Submission(s): 5982
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
直接三個循環必定會超時,所以用二分算法即可,a+b各種情況算出來,然後x-c二分查找就行了。
看了半天,原來j寫成i了。
代碼如下:
#include <cstdio>
#include <cstring>
#include <math.h>
#include <algorithm>
#include <iostream>
using namespace std;
int l,m,n,t,flag,sum[510*510],a[510],b[510],c[510];
void cmp(int x)
{
int mid,z=0,y=t-1;
while(z<=y)
{
mid=(z+y)>>1;
if(x>sum[mid]) z=mid+1;
else if(x<sum[mid]) y=mid-1;
else {flag=1;return ;}
}
return ;
}
int main()
{
int i,j,cnt=1,x,s;
while(scanf("%d%d%d",&l,&m,&n)!=EOF)
{
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]);
t=0;
for(i=0;i<l;i++)
for(j=0;j<m;j++)
sum[t++]=a[i]+b[j];
sort(sum,sum+t);
scanf("%d",&s);
printf("Case %d:\n",cnt++);
while(s--)
{
scanf("%d",&x);
flag=0;
for(i=0;i<n;i++)
{
cmp(x-c[i]);
if(flag)
{
printf("YES\n");
break;
}
}
if(!flag) printf("NO\n");
}
}
return 0;
}