杭电—2141

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

运用二分的思想;

#include<stdio.h>

#include<stdlib.h>

#include<iostream>

#include<algorithm>

using namespace std;

#define maxn 510

int a[maxn],b[maxn],c[maxn];

int sum[maxn*maxn];

int bin_search(int y,int cnt)

{

    int left=0,right=cnt-1;

    int mid;

    while(left<=right)

    {

        mid=(left+right)/2;

        if(sum[mid]==y) return 1;

        if(sum[mid]<y) left= mid+1;

        if(sum[mid]>y) right=mid-1;

    }

    return 0;

}

int main ()

{

    int l,m,n;

    int k=1;

    while(~scanf(“%d%d%d”,&l,&m,&n))

    {

        for(int i=0; i<l; i++)

            scanf(“%d”,&a[i]);

        for(int i=0; i<m; i++)

            scanf(“%d”,&b[i]);

        for(int i=0; i<n; i++)

            scanf(“%d”,&c[i]);

        int cnt=0;

        for(int i=0; i<l; i++ )

            for(int j=0; j<m; j++)

            {

                sum[cnt++]=a[i]+b[j];

            }

        sort(sum,sum+cnt);\\sort是一个排序函数,可以自己百度了解一下

        int s;

        scanf(“%d”,&s);

        printf(“Case %d:\n”,k++);

        while(s–)

        {

            int x;

            scanf(“%d”,&x);

            int flag=0;

            for(int i=0; i<n; i++)

            {

                if(bin_search(x-c[i],cnt))

                {

                    flag=1;

                    break;

                }

            }

            if(flag==1)printf(“YES\n”);

            else printf(“NO\n”);

        }

    }

    return 0;

}

\\Ai+Bj+Ck = X可以写成Ai+Bj=X-Ck;

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