HDU 2141 Can you find it?【二分查找是否存在ai+bj+ck=x】

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

【STL版本】

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,x,n) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long ll;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const ll LNF = 1e18;
const int maxn = 1e3 + 20;
const int maxm = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
ll quickpow(ll a, ll b) {
    ll ans = 0;
    while (b > 0) {
        if (b % 2)ans = ans * a;
        b = b / 2;
        a = a * a;
    }
    return ans;
}
int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a%b);
}
bool cmp(int a, int b) {
    return a > b;
}
int l,n,m,q,x;
int a[maxn],b[maxn],c[maxn];
int ab[250005];
/*
ai + bj + ck = x ——>
ai + bj = x -ck
1 2 3
1 2 3
1 2 3
*/
int main()
{
    int cas = 1;
    while(~scanf("%d%d%d",&l,&n,&m))
    {
        int f = 0, k;
        ms(a,0),ms(b,0),ms(c,0),ms(ab,0);
        rep(i,0,l)
            scanf("%d",&a[i]);
        rep(i,0,n)
            scanf("%d",&b[i]);
        rep(i,0,m)
            scanf("%d",&c[i]);
        k = 0;
        rep(i,0,l)
        {
            rep(j,0,n)
            {
                ab[k++] = a[i] + b[j];
            }
        }
        sort(ab,ab+k);
        sort(c,c+m);
        printf("Case %d:\n",cas++);
        scanf("%d",&q);
        while(q--)
        {
            //在ab数组二分查找 x - c[i]
            f = 0;
            scanf("%d",&x);
            for(int j=0;j<m;j++)
            {
                int pos = lower_bound(ab,ab+k,x-c[j]) - ab;
                if(ab[pos] == x - c[j])
                {
                    f = 1;
                    break;
                }
            }
            if(f) printf("YES\n");
            else printf("NO\n");
        }
    }
}

【手写二分版本】：

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,x,n) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long ll;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const ll LNF = 1e18;
const int maxn = 1e3 + 20;
const int maxm = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
ll quickpow(ll a, ll b) {
    ll ans = 0;
    while (b > 0) {
        if (b % 2)ans = ans * a;
        b = b / 2;
        a = a * a;
    }
    return ans;
}
int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a%b);
}
bool cmp(int a, int b) {
    return a > b;
}
int l,n,m,q,x;
int a[maxn],b[maxn],c[maxn];
int ab[250005];
/*
ai + bj + ck = x ——>
ai + bj = x -ck
1 2 3
1 2 3
1 2 3
*/
int main()
{
    int cas = 1;
    while(~scanf("%d%d%d",&l,&n,&m))
    {
        int f = 0, k;
        ms(a,0),ms(b,0),ms(c,0),ms(ab,0);
        rep(i,0,l)
            scanf("%d",&a[i]);
        rep(i,0,n)
            scanf("%d",&b[i]);
        rep(i,0,m)
            scanf("%d",&c[i]);
        k = 0;
        rep(i,0,l)
        {
            rep(j,0,n)
            {
                ab[k++] = a[i] + b[j];
            }
        }
        sort(ab,ab+k);
        sort(c,c+m);
        printf("Case %d:\n",cas++);
        scanf("%d",&q);
        while(q--)
        {
            //在ab数组二分查找 x - c[i]
            f = 0;
            scanf("%d",&x);
            for(int j=0;j<m;j++)
            {
                int l = 0, r = k - 1, mid;
                while(l <= r)
                {
                    mid = (l+r)/2;
                    if(ab[mid] == x-c[j])
                    {
                        f=1;break;
                    }
                    else if(ab[mid]<x-c[j]) l=mid+1;
                    else if(ab[mid]>x-c[j]) r=mid-1;
                }
                if(f) break;
            }
            if(f) printf("YES\n");
            else printf("NO\n");
        }
    }
}