A - Gaby And Addition
题意：给n个数，现在规定一个操作是不进位的加法，问两两操作的最小值和最大值。
思路：既然是不进位，那每个位之间就没有影响了。我们现在想要两两操作的值最大的话，就相当于找其他n-1个数，跟他们的前缀匹配一下，找出每一位能产生的最大值((cur+bit[i])%10的最大值)。这样就可以用字典树维护前面的数的前缀，每次只用匹配一下往最值方向走就行了。最小值同理。详见代码。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '
'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e6+200;
const ll inf=0x3f3f3f3f3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
ll n;
ll trie[13*maxn][10];
ll a[maxn];
ll pw[20];
int k = 1;
inline void insert(ll x)
{
int p = 0;
vector<int> tmp;
rep(i,0,18) tmp.pb(x%10), x /= 10;
per(i,18,0)
{
int ch = tmp[i];
if(!trie[p][ch]) trie[p][ch] = k++;
p = trie[p][ch];
}
}
inline ll search(ll x, bool flag)
{
int p = 0;
vector<int> tmp;
rep(i,0,18) tmp.pb(x%10), x /= 10;
ll ans = 0;
per(i,18,0)
{
int ch = tmp[i];
if(flag)
{
int maxj = -1, cur = -1;
per(j,9,0) if(trie[p][j]&&(ch + j)%10 > cur) cur = (ch+j)%10, maxj = j;
p = trie[p][maxj], ans = ans + pw[i]*((ch + maxj) % 10);
}
else
{
int minj = -1, cur = 11;
per(j,9,0) if(trie[p][j]&&(ch + j)%10 < cur) cur = (ch+j)%10, minj = j;
p = trie[p][minj], ans = ans + pw[i]* ((ch + minj) % 10);
}
}
return ans;
}
signed main()
{
pw[0] = 1; ll ans1 = inf, ans2 = 0;
for(int i = 1; i <= 18; i++) pw[i] = pw[i-1] * 10;
n = read();
rep(i,1,n)
{
a[i] = read();
if(i>1)
ans1 = min(ans1,search(a[i],false)) ,ans2 = max(ans2, search(a[i],true));
insert(a[i]);
}
cout<<ans1<<' '<<ans2<<'
';
return 0;
}
B - Maximum Tree
题意：给n个数，随意排列，使得在a[k]代表树第k层的结点所含有的子树个数的前提下，树的结点个数最大值。
贪心，把子树个数多的放到靠近根的地方即可。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '
'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
ll a[maxn];
int main()
{
ll n = read();
rep(i,1,n) a[i] = read();
sort(a+1,a+1+n);
ll ans = 1;
ll cur = 1;
ll pre = 1;
per(i,n,1)
{
cur = pre*a[i];
ans += cur;
pre = cur;
}
cout<<ans<<endl;
return 0;
}
C - Planet Communcation
题意：问最少要多少条过原点的直线可以覆盖图中所有点。
思路：以第一个点作为原点，然后构建向量，化成最简，然后看向量种类有多少种即是答案。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '
'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
typedef struct Pos
{
ll x;
ll y;
ll z;
// bool operator < (Pos a)const{return x<a.x;}
}P;
P a[maxn];
map<pair<PII,ll>,ll> cnt;
int main()
{
ll n = read();
a[1].x = read(), a[1].y = read(), a[1].z = read();
rep(i,2,n)
{
a[i].x = read(), a[i].y = read(), a[i].z = read();
a[i].x -= a[1].x, a[i].y -= a[1].y, a[i].z -= a[1].z;
ll d = gcd(a[i].x, a[i].y);
d = gcd(d, a[i].z);
P tmp;
tmp.x = a[i].x/d, tmp.y = a[i].y/d, tmp.z = a[i].z/d;
cnt[mp(mp(tmp.x, tmp.y),tmp.z)]++;
}
cout<<cnt.size()<<endl;
return 0;
}
D - Double it
题意：给一个操作A:x->2x+1, B:x->2x+2，一开始x为0，问到n的方案。
思路：正着推发现有难度。但是发现到A可以唯一确定答案为一个奇数，B可以唯一确定答案为偶数。所以如果n一开始是奇数，那么肯定上一步是A过来的，偶数的话就肯定是B过来的，然后继续往回推。最后倒序输出方案就完事了。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '
'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
int main()
{
ll n;
n = read();
string s;
while(n)
{
if(n&1) n--, n/=2, s += 'A';
else n -= 2, n /= 2, s+= 'B';
}
per(i,s.size()-1,0) cout<<s[i]; cout<<endl;
return 0;
}
E - Text Editor
题意：给一个主串t和模式串p，问在p的最长的且在t中出现次数超过k次的前缀。
思路：首先会想到KMP。但是如果直接拿模式串一位一位匹配是O(n2)的时间复杂度，肯定过不了。
我们发现模式串长度越长，越不容易满足要求。所以这里有个单调性，可以来二分。我们二分答案可能的长度，然后用这么长的前缀丢进KMP里面，看看能否满足，如果发现可以，就把长度调大一点，反之调小一点。时间复杂度O(nlogn)。
最后注意KMP用来计数时，每次匹配完不能直接j=0, i -= p.size()，因为这样会使得指针回溯使得算法退化。我们匹配完一个直接j = nxt[j]就行了（还是尽量选择前面最大匹配完的一部分接着匹配）。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '
'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
int nxt[maxn];
void getnext(string p)
{
mem(nxt,0);
nxt[0] = -1; //第一位设为-1
int i = 0, j = -1; //i指针用来遍历数组， j相当于最大公共前后缀长度
while(i<(int)p.size())
{
if(j==-1||p[i]==p[j]) i++, j++, nxt[i] = j; //看看前后缀是否相等
else j = nxt[j]; // 不行的话往前回溯
}
}
int KMP(string t, string p)
{
int i = 0, j = 0;
getnext(p);
int cnt = 0;
while(i<t.size())
{
if(j==p.size())
{
cnt++;
j = nxt[j];
}
if(j==-1||t[i]==p[j]) i++, j++; //相等继续
else j = nxt[j]; // 失配后找到nxt[j]的位置继续匹配。
}
if(j==p.size()) cnt++;
return cnt;
}
int main()
{
string t, p;
int n;
getline(cin,t);
getline(cin,p);
cin>>n;
int L = 1 , R = p.size();
string res = "-1";
while(L<=R)
{
int mid = (L+R)>>1;
string tmp(p,0, mid);
if(KMP(t,tmp)>=n) res = tmp, L = mid + 1;
else R = mid - 1;
}
if(res=="-1") cout<<"IMPOSSIBLE"<<endl;
else cout<<res<<endl;
return 0;
}
F - Polygon Triangles
题意：给你n个三角形，问你能不能够组成一个非退化四边形。
思路：发现其实是个三角形就能满足。。。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '
'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
typedef struct Pos
{
ll x;
ll y;
ll z;
}P;
P a[maxn];
map<pair<PII,ll>,ll> cnt;
bool check(ll a, ll b, ll c)
{
if(a+b>c&&a-b<c) return true;
return false;
}
int main()
{
ll n = read();
bool flag = 1;
rep(i,1,n)
{
ll b[5];
b[1] = read(), b[2] = read(), b[3] = read();
sort(b+1,b+1+3);
a[i].x = b[1] , a[i].y = b[2], a[i].z = b[3];
if(!check(a[i].x, b[2], b[3])) flag = 0;
if(flag)
cnt[mp(mp(a[i].x,a[i].y),a[i].z)]++;
}
if(flag) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
return 0;
}
H - Logo
题意：输出指定规模的UN。
思路：水题模拟。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '
'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
ll a[maxn];
int main()
{
ll n = read();
cout<<"*";
rep(i,1,n-2) cout<<' '; cout<<"* ";
rep(i,1,n) cout<<"*"; cout<<endl;
rep(i,1,n-2)
{
cout<<"*";
rep(j,1,n-2) cout<<' ';
cout<<"* ";
cout<<"*";
rep(j,1,n-2) cout<<' '; cout<<"*"; cout<<endl;
}
rep(i,1,n) cout<<"*";
cout<<" *";
rep(i,1,n-2) cout<<' '; cout<<"*";cout<<endl;
return 0;
}
J - Jeronimo's List
题意：给出n个数，q个询问，每次问你第几个数是什么（从小到大） , n <= 1e7。
思路：直接sort肯定白给。我们可以用桶排序的思想，先预处理好这个完整的数组，然后把每个值映射到1e7的范围里，统计好个数，然后从头到尾走一遍，先遇到的就是靠前的。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '
'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 3e7+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 3e7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
ll a[maxn];
ll ans[maxn];
ll id[maxn];
int main()
{
ll n = read(), m = read(), q = read();
rep(i,1,m) a[i] = read();
rep(i,m+1,n) a[i] = (a[i-m]+a[i-m+1])%mod;
rep(i,1,n) ans[a[i]]++;
ll cur = 1;
int p = 0;
rep(i,0,mod-1)
{
if(ans[i])
{
while(ans[i])
id[++p] = i, cur ++, ans[i]--;
}
}
rep(i,1,q)
{
ll x = read();
cout<<id[x]<<endl;
}
return 0;
}