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");
}
}
}