hdu2141 (二分）

来源：互联网发布：java用core生成二维码编辑：程序博客网时间：2024/05/03 23:15

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

Author

wangye

Source

HDU 2007-11 Programming Contest

题意：分别有a,b,c三个数组，给出一个数，看这个数能不能再a,b,c,里面分别找出一个数求和得到

思路：如果暴力的话肯定会超时；我们可以将三个数组规划成两个数组，其中一个数组是两个数组的和，并且排序去重。对于每个给出的数，我们依次枚举，然后二分搜索出是否有结果。

代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;const int N=505;int len;int a[N],b[N],c[N],d[N*N];int find(int x){    int l=0,r=len-1;    while(l<r)    {        int temp=(l+r)/2;        if(d[temp]<x)            l=temp+1;        else            r=temp;    }    if(x==d[l])        return 1;    else        return 0;}int main(){    int l,n,m,t=1;    while(~scanf("%d%d%d",&l,&n,&m))    {        for(int i=0;i<l;i++)            scanf("%d",&a[i]);        sort(a,a+l);        for(int i=0;i<n;i++)            scanf("%d",&b[i]);        for(int i=0;i<m;i++)            scanf("%d",&c[i]);        int s;        scanf("%d",&s);        printf("Case %d:\n",t++);        len=0;        for(int i=0;i<n;i++)            for(int j=0;j<m;j++)                d[len++]=b[i]+c[j];        sort(d,d+len);        len=unique(d,d+len)-d;        for(int i=1;i<=s;i++)        {            int x,flag=0;            scanf("%d",&x);            for(int j=0;j<l;j++)            {                int y=x-a[j];                if(find(y))                {                    flag=1;                    printf("YES\n");                    break;                }            }            if(!flag)                printf("NO\n");        }    }}

0 0