[考试总结]noip模拟69(多校)
石子游戏
这个的部分分数 \(50pts\) 需要使用 \(SG\) 函数。。
因为上次颓废，然后就没有学。。。
一个堆的 \(SG\) 函数就是最高的那个 \(SG\)
然后我们的 \(SG\) 函数就是递推而成。
我们将这个东西所能到达的所有状态的 \(SG\) 值排序，之后我们就可以找到第一个没出现的非负整数，这个就是他的 \(SG\) 值。
然后这样的话是 \(\mathcal O(n^3)\) 的，只有 \(20pts\)
有一个规律，就是一个堆的 \(SG\) 值就是这个 \(a_i \% (k+1)\)，之后就可以 \(50pts\) 了。
#include<bits/stdc++.h>
using namespace std;
#define try(i,a,b) for(register int i=a;i<=b;++i)
#define throw(i,a,b) for(register int i=a;i>=b;--i)
#define go(i,x) for(register signed i=head[x],y=edge[i].ver;i;i=edge[i].next,y=edge[i].ver)
namespace xin_io
{
#define file(x) FILE *FI = freopen(#x".in","r",stdin); FI = freopen(#x".out","w",stdout);
#define sb(x) cout<<#x" = "<<x<<' '
#define jb(x) cout<<#x" = "<<x<<endl
#define debug cout<<"debug"<<endl
#define gec() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1 ++
#define gc() getchar()
#define scanf ak = scanf
char buf[1<<20],*p1 = buf,*p2 = buf; using ll = long long; using ull = unsigned long long; int ak;
class xin_stream{public:template<typename type>xin_stream &operator >> (type &s)
{
s = 0; register bool f = 0; register char ch = gc();
while(!isdigit(ch)) f |= ch == '-',ch = gc();
while( isdigit(ch)) s = (s << 1) + (s << 3) + (ch xor 48),ch = gc(); return s = f ? -s : s,*this;
}}io;
}
using namespace xin_io; static const int maxn = 1e6+10,two = 4e3+10,inf = 1e9+10;
#define int long long
namespace xin
{
int sg[maxn];
int n,a[maxn],maxx;
int temp[maxn],zhi = 0;
bool vis[maxn];
inline short main()
{
file(stone);
io >> n;
try(i,1,n) io >> a[i],maxx = max(maxx,a[i]);
auto check = [](int k) -> bool
{
// try(i,1,maxx)
// {
// zhi = 0;
// throw(j,i-1,i-k)
// if(j < 0) break;
// else temp[++zhi] = sg[j],vis[sg[j]] = 1;
//// std::sort(temp+1,temp+zhi+1);
// for(register int j=0;;j++) if(!vis[j]) {sg[i] = j; break;}
// try(j,1,zhi) vis[temp[j]] = 0;
// }
int ret = 0;
try(i,1,n) ret ^= (a[i] % (k + 1));
// jb(k);
// try(i,1,maxx) jb(sg[i]);
return ret;
};
try(i,1,n)
printf(check(i) ? "Alice " : "Bob ");
return 0;
}
}
signed main() {return xin::main();}
黑客1
这个就是垃圾题。
枚举就行了。
然后将两个倍数的区间进行判断就行，其实就是找到交集。。。
#include<bits/stdc++.h>
using std::cout; using std::endl;
#define try(i,a,b) for(register int i=a;i<=b;++i)
#define throw(i,a,b) for(register int i=a;i>=b;--i)
#define go(i,x) for(register signed i=head[x],y=edge[i].ver;i;i=edge[i].next,y=edge[i].ver)
namespace xin_io
{
#define file(x) FILE *FI = freopen(#x".in","r",stdin); FI = freopen(#x".out","w",stdout);
#define sb(x) cout<<#x" = "<<x<<' '
#define jb(x) cout<<#x" = "<<x<<endl
#define debug cout<<"debug"<<endl
#define gec() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1 ++
#define gc() getchar()
#define scanf ak = scanf
char buf[1<<20],*p1 = buf,*p2 = buf; using ll = long long; using ull = unsigned long long; int ak;
class xin_stream{public:template<typename type>xin_stream &operator >> (type &s)
{
s = 0; register bool f = 0; register char ch = gc();
while(!isdigit(ch)) f |= ch == '-',ch = gc();
while( isdigit(ch)) s = (s << 1) + (s << 3) + (ch xor 48),ch = gc(); return s = f ? -s : s,*this;
}}io;
}
using namespace xin_io; static const int maxn = 1e6+10,two = 4e3+10,inf = 1e9+10;
#define int long long
namespace xin
{
const int mod = 1e9+7;
int a,b,c,d,jsq,ans;
inline int gcd(int x,int y) {return !y ? x : gcd(y,x%y);}
inline short main()
{
freopen("hacker.in","r",stdin); freopen("hacker.out","w",stdout);
io >> a >> b >> c >> d;
if(b <= 1000 and d <= 1000)
{
auto check = [](int x,int y)
{
register int yue = gcd(x,y);
if(yue) x /= yue,y /= yue;
if(x + y <= 999) return x+y;
return 0ll;
};
try(i,a,b) try(j,c,d)
jsq += check(i,j),jsq -= (jsq >= mod) ? mod : 0;
cout<<jsq % mod<<endl;
return 0;
}
int ms1 = std::min(999ll,b),ms2 = std::min(999ll,d);
try(i,1,ms1) try(j,1,ms2-i)
if(gcd(i,j) == 1)
{
int l1,r1,l2,r2;
l1 = (a + i - 1) / i; r1 = b / i;
l2 = (c + j - 1) / j; r2 = d / j;
if(r1 >= l2 and l1 <= l2 and r2 >= r1)
ans += (r1 - l2 + 1) * (i + j) % mod,ans %= mod;
else if(r2 >= l1 and l2 <= l1 and r1 >= r2)
ans += (r2 - l1 + 1) * (i + j) % mod,ans %= mod;
else if(r2 >= r1 and l2 <= l1) ans += (r1 - l1 + 1) * (i + j) % mod,ans %= mod;
else if(r1 >= r2 and l2 >= l1) ans += (r2 - l2 + 1) * (i + j) % mod,ans %= mod;
}
cout<<ans<<endl;
return 0;
}
}
signed main() {return xin::main();}
黑客2
垃圾高精，超级卡常，然后使用 \(nb\) 方法可以过掉。。
#include<bits/stdc++.h>
using namespace std;
#define try(i,a,b) for(register int i=a;i<=b;++i)
#define throw(i,a,b) for(register int i=a;i>=b;--i)
#define go(i,x) for(register signed i=head[x],y=edge[i].ver;i;i=edge[i].next,y=edge[i].ver)
namespace xin_io
{
#define file(x) FILE *FI = freopen(#x".in","r",stdin); FI = freopen(#x".out","w",stdout);
#define sb(x) cout<<#x" = "<<x<<' '
#define jb(x) cout<<#x" = "<<x<<endl
#define debug cout<<"debug"<<endl
#define gec() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1 ++
#define gc() getchar()
#define scanf ak = scanf
char buf[1<<20],*p1 = buf,*p2 = buf; using ll = long long; using ull = unsigned long long; int ak;
class xin_stream{public:template<typename type>xin_stream &operator >> (type &s)
{
s = 0; register bool f = 0; register char ch = gc();
while(!isdigit(ch)) f |= ch == '-',ch = gc();
while( isdigit(ch)) s = (s << 1) + (s << 3) + (ch xor 48),ch = gc(); return s = f ? -s : s,*this;
}}io;
}
using namespace xin_io; static const int maxn = 1e6+10,two = 4e3+10,inf = 1e9+10;
int n,m,k;
#define int long long
namespace xin
{
const ll bse = 1e9;
struct bigint
{
ll a[250];int len;
bigint(){len = 1;}
bigint(int x){
memset(a,0,sizeof(a));len = 1;
while(x){a[len++] = x % bse,x /= bse;}len --;
}
void print(){
printf("%lld",a[len]);
for(register int i = len-1 ; i ; i --)printf("%09lld",a[i]);
}
void pprint()
{
printf("%lld",a[len]);
throw(i,len-1,1) printf("%017lld",a[i]);
// printf("%016lld",a[1]);
}
friend bigint operator + (const bigint& a,const bigint& b){
bigint ret = 0; ret.len = max(a.len,b.len) + 1;
for(register int i = 1 ; i <= ret.len ; i ++)
ret.a[i] += a.a[i] + b.a[i],
ret.a[i+1] += ret.a[i] / bse,
ret.a[i] %= bse;
for(register int i = 1 ; i <= ret.len ; i ++)
ret.a[i+1] += ret.a[i] / bse,
ret.a[i] %= bse;
while(ret.len > 1 && !ret.a[ret.len]) ret.len --;
return ret;
}
friend bigint operator * (const bigint& a,const bigint& b){
bigint ret = 0; ret.len = a.len + b.len + 1;
for(register int i = 1 ; i <= a.len ; i ++)
for(int j = 1 ; j <= b.len ; j ++)
ret.a[i+j-1] += a.a[i] * b.a[j],
ret.a[i+j] += ret.a[i+j-1] / bse,
ret.a[i+j-1] %= bse;
for(register int i = 1 ; i <= ret.len ; i ++)
ret.a[i+1] += ret.a[i] / bse ,
ret.a[i] %= bse;
while(ret.len > 1 && !ret.a[ret.len]) ret.len --;
return ret;
}
friend bigint operator * (const bigint& a,const int &b){
bigint ret = 0 ; ret.len = a.len;
int x = b;
while(x) ret.len ++ , x /= bse;
for(register int i = 1 ; i <= a.len ; i ++)
ret.a[i] += a.a[i] * b,
ret.a[i+1] += ret.a[i] / bse,
ret.a[i] %= bse;
for(register int i = 1 ; i <= ret.len ; i ++)
ret.a[i+1] += ret.a[i] / bse,
ret.a[i] %= bse;
while(ret.len > 1 && !ret.a[ret.len]) ret.len --;
return ret;
}
};
bigint ans1 = 0,ans2 = 0,ans = 0,pow2 = 1;
bool vis[two][two];
int st[maxn];
bigint f[130001];
class xin_data
{
public:
int x,y;
xin_data(){}
xin_data(int x,int y):x(x),y(y){}
}d[two/5][two/5];
int tot;
bigint shi[501];
xin_data dfs(int x,int s)
{
if(vis[x][s]) return d[x][s];
vis[x][s] = 1;
if(x == n + 1) return d[x][s] = xin_data(1,0);
register int temp1 = ++tot,temp2 = ++tot;
try(i,1,k) if(!(s & (1 << i - 1)))
{
xin_data temp = dfs(x+1,s|st[i]);
f[temp1] = f[temp1] + f[temp.x];
f[temp2] = f[temp2] + f[temp.y];
bigint wk = f[temp.x] * i * shi[n-x];
f[temp2] = f[temp2] + wk;
}
return d[x][s] = xin_data(temp1,temp2);
}
inline short main()
{
try(i,1,m)
{
register int x,y; io >> x >> y;
st[x] |= (1 << y - 1);
}
shi[0] = 1;
try(i,1,500) shi[i] = shi[i-1] * 10;
f[1] = 1; tot = 1;
xin_data ans = dfs(1,0);
f[ans.x].print(); cout<<endl;
f[ans.y].print(); cout<<endl;
return 0;
}
}
#undef int
using namespace std;
namespace xin1
{
const int maxn = 10100;
struct bigint
{
int d[maxn], len;
void clean() { while(len > 1 && !d[len-1]) len--; }
bigint() { memset(d, 0, sizeof(d)); len = 1; }
bigint(int num) { *this = num; }
bigint(char* num) { *this = num; }
bigint operator = (const char* num)
{
memset(d, 0, sizeof(d)); len = strlen(num);
for(int i = 0; i < len; i++) d[i] = num[len-1-i] - '0';
clean();
return *this;
}
bigint operator = (int num)
{
char s[21]; sprintf(s, "%d", num);
*this = s;
return *this;
}
bigint operator + (const bigint& b){
bigint c = *this; int i;
for (i = 0; i < b.len; i++){
c.d[i] += b.d[i];
if (c.d[i] > 9) c.d[i]%=10, c.d[i+1]++;
}
while (c.d[i] > 9) c.d[i++]%=10, c.d[i]++;
c.len = max(len, b.len);
if (c.d[i] && c.len <= i) c.len = i+1;
return c;
}
bigint operator - (const bigint& b){
bigint c = *this; int i;
for (i = 0; i < b.len; i++){
c.d[i] -= b.d[i];
if (c.d[i] < 0) c.d[i]+=10, c.d[i+1]--;
}
while (c.d[i] < 0) c.d[i++]+=10, c.d[i]--;
c.clean();
return c;
}
bigint operator * (const bigint& b)const{
int i, j; bigint c; c.len = len + b.len;
for(j = 0; j < b.len; j++) for(i = 0; i < len; i++)
c.d[i+j] += d[i] * b.d[j];
for(i = 0; i < c.len-1; i++)
c.d[i+1] += c.d[i]/10, c.d[i] %= 10;
c.clean();
return c;
}
bigint operator / (const bigint& b){
int i, j;
bigint c = *this, a = 0;
for (i = len - 1; i >= 0; i--)
{
a = a*10 + d[i];
for (j = 0; j < 10; j++) if (a < b*(j+1)) break;
c.d[i] = j;
a = a - b*j;
}
c.clean();
return c;
}
bool operator <(const bigint& b) const{
if(len != b.len) return len < b.len;
for(int i = len-1; i >= 0; i--)
if(d[i] != b.d[i]) return d[i] < b.d[i];
return false;
}
bool operator >(const bigint& b) const{return b < *this;}
std::string str() const{
char s[maxn]={};
for(int i = 0; i < len; i++) s[len-1-i] = d[i]+'0';
return s;
}
};
istream& operator >> (istream& in, bigint& x)
{
std::string s;
in >> s;
x = s.c_str();
return in;
}
ostream& operator << (ostream& out, const bigint& x)
{
out << x.str();
return out;
}
inline short main()
{
bigint ans = 0;
try(i,1,n) ans = ans * 10 + 1;
ans = ans * (k + 1) * 10;
ans = ans / 2;
bigint pow2 = 1;
try(i,1,n) pow2 = pow2 * k;
cout<<pow2<<endl;
ans = ans * pow2;
cout<<ans / 10<<endl;
return 0;
}
}
signed main()
{
file(hacker2);
io >> n >> m >> k;
if(!m) return xin1::main(),0;
else return xin::main(),0;
}
