HDU - 2141 二分法

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

題意描述：這道題題意很清晰，即給你三個數組A,B,C是否等於給定的數值X
Ai+Bj+Ck=X;
暴力枚舉會炸，可以先處理前兩個數字的和O(n^2),排序，之後枚舉X-Ck,二分找

二分查找

步驟1、設置數組s[n]來存放n個已排好序的元素；變量low和high表示查找範圍在數組中的上界和下界；middle表示查找範圍的中間位置，x爲特定元素

步驟2、初始化，low=0，即指向s中的第一個元素；high=n-1，即指向s中的最後一個元素

步驟3:、middle=（low+high）/2，即指向查找範圍的中間元素

步驟4、判定low<=high是否成立，如果成立，轉步驟5，否則，算法結束

步驟5、判斷x與s[middle] 的關係，如果等於s[middle],算法結束，如果大於s[middle]，則令low=middle+1;否則令high=middle-1，轉步驟3

代碼實現

//非遞歸方式
int dd(int a[],int n,int x){//x表示要查找的元素
int low=0,high=n-1;
    int mid;
    while(low<=high){
        mid=(low+high)/2;
        if(a[mid]>x){
            high=mid-1;
        }
        else if(a[mid]<x){
            low=mid+1;
        }
        else return mid;
    }
    return -1;
}
//遞歸方式
int dd(int s[],int x,int low,int high){
if(low>high)
return -1;
int mid=(low+high)/2;
if(x==s[mid])
return mid;
else if(x>s[mid])
return dd(s,x,mid+1,high);
else
return dd(s,x,low,mid-1);
}

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

using namespace std;
/****二分法   暴力枚舉(三個for)會超時
計算每個數組找一個數之和是否能夠等於給定 的數值****/
int l[505],n[505],m[505];
int a[250005];//l與n的組合
int dd(int low,int high,int x){
    int mid;
    while(low<=high){
        mid=(low+high)/2;
        if(a[mid]>x){
            high=mid-1;
        }
        else if(a[mid]<x){
            low=mid+1;
        }
        else return 1;
    }
    return -1;
}
int main(){
int L,N,M;
int len,t;
int S;
int num=0;
int x,k;
while(scanf("%d%d%d",&L,&N,&M)>0){
    for(int i=1;i<=L;i++){
        scanf("%d",&l[i]);
    }
    for(int i=1;i<=N;i++){
        scanf("%d",&n[i]);
    }
    for(int i=1;i<=M;i++){
        scanf("%d",&m[i]);
    }
    len=L*N;
    t=0;
    for(int i=1;i<=L;i++){
        for(int j=1;j<=N;j++){
            a[++t]=l[i]+n[j];
        }
    }
    //排序
   sort(a+1,a+1+len);
    scanf("%d",&S);
    printf("Case %d:\n",++num);
    while(S--){
   scanf("%d",&x);
            for(int i=1;i<=M;i++)
            {
                k=dd(1,len,x-m[i]);
                 if( k==1 ) break;
            }
            if( k==-1) printf("NO\n");
            else printf("YES\n");
    }
}
return 0;
}