Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 10424    Accepted Submission(s): 2742

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 31 2 31 2 31 2 331410

 

Sample Output

Case 1:NOYESNO

 

第一次二分，，尼玛刚开始写的很挫啊，，不过现在还好。。

#include<cstdio>#include<algorithm>using namespace std;int a[505],b[505],c[505],f[250010];int binary(int x,int M,int k){    for(int i=0; i<M; i++)    {        int l = 0, r = k;        while(l <= r)        {            int mid = (l+r)/2;            if(x > c[i]+f[mid]) l = mid+1;            else if(x < c[i]+f[mid]) r = mid-1;            else return 1;        }    }    return 0;}int main(){    int L,N,M,i,j,s,t=1;    while(~scanf("%d%d%d", &L,&N,&M))    {        for(i=0; i<L; i++)            scanf("%d", a+i);        for(i=0; i<N; i++)            scanf("%d", b+i);        for(i=0; i<M; i++)            scanf("%d", c+i);        int k = 0;        for(i=0; i<L; i++)            for(j=0; j<N; j++)                f[k++] = a[i]+b[j];        sort(f,f+k);        scanf("%d", &s);        printf("Case %d:\n", t++);        while(s--)        {            int x;            scanf("%d", &x);            printf(binary(x,M,k-1) ? "YES\n" : "NO\n");        }    }    return 0;}