Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 22312 Accepted Submission(s): 5627
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
Author
wangye
题目大意：给你三个数列，然后给你若干个sum,问你能否从每个数列中各取出一个数，使得它们的和为sum，如果可以，输出YES,
反之输出NO。
思路分析：很明显暴力做会超时，应该采用先合并数组然后二分查找，时间复杂度降到了nlogn，但是在实现的时候还是有几点要注意，
首先是输出格式，先 输出Case，然后再输入s，另外由于粗心大意，low和high的声明写到了for循环外面，wa到死，沧桑。
代码：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
const int inf=0xffffff;
const int maxn=500+50;
__int64 a[maxn],b[maxn],c[maxn];
__int64 f[maxn*maxn];
int kase=0;
int main()
{
int L,N,M;
__int64 s,sum;
while(scanf("%d%d%d",&L,&N,&M)!=EOF)
{
int num=0;
for(int i=0;i<L;i++)
scanf("%I64d",&a[i]);
for(int i=0;i<N;i++)
scanf("%I64d",&b[i]);
for(int i=0;i<M;i++)
scanf("%I64d",&c[i]);
for(int i=0;i<L;i++)
{
for(int j=0;j<N;j++)
{
f[num++]=a[i]+b[j];
}
}
sort(f,f+num);
printf("Case %d: ",++kase);
scanf("%I64d",&s);
while(s--)
{
scanf("%I64d",&sum);
int flag=0;
for(int i=0;i<M;i++)
{
int low=0,high=L*N-1;
if(flag) break;
while(low<=high)
{
int mid=(high+low)/2;
if(f[mid]==sum-c[i])
{
flag=1;
break;
}
else if(f[mid]<sum-c[i]) low=mid+1;
else high=mid-1;
}
}
if(flag) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}
return 0;
}
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
const int inf=0xffffff;
const int maxn=500+50;
__int64 a[maxn],b[maxn],c[maxn];
__int64 f[maxn*maxn];
int kase=0;
int main()
{
int L,N,M;
__int64 s,sum;
while(scanf("%d%d%d",&L,&N,&M)!=EOF)
{
int num=0;
for(int i=0;i<L;i++)
scanf("%I64d",&a[i]);
for(int i=0;i<N;i++)
scanf("%I64d",&b[i]);
for(int i=0;i<M;i++)
scanf("%I64d",&c[i]);
for(int i=0;i<L;i++)
{
for(int j=0;j<N;j++)
{
f[num++]=a[i]+b[j];
}
}
sort(f,f+num);
printf("Case %d: ",++kase);
scanf("%I64d",&s);
while(s--)
{
scanf("%I64d",&sum);
int flag=0;
for(int i=0;i<M;i++)
{
int low=0,high=L*N-1;
if(flag) break;
while(low<=high)
{
int mid=(high+low)/2;
if(f[mid]==sum-c[i])
{
flag=1;
break;
}
else if(f[mid]<sum-c[i]) low=mid+1;
else high=mid-1;
}
}
if(flag) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}
return 0;
}
tip:直接调用二分查找函数binary_search也可以解决