A. ASCII Addition
模拟
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(_i,_a,_n) for(int _i=_a;_i<=_n;++_i)
#define PER(_i,_a,_n) for(int _i=_n;_i>=_a;--_i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(_a) ({REP(_i,1,n) cout<<_a[_i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
const int N = 1e6+10;
char s[10][N],ans[10][N];
int id(int i) {
if (string(s[1]+i,s[1]+i+5)=="....x") {
if (string(s[2]+i,s[2]+i+5)=="....x")
if (string(s[3]+i,s[3]+i+5)=="....x")
if (string(s[4]+i,s[4]+i+5)=="....x")
if (string(s[5]+i,s[5]+i+5)=="....x")
if (string(s[6]+i,s[6]+i+5)=="....x")
if (string(s[7]+i,s[7]+i+5)=="....x")
return 1;
}
if (string(s[1]+i,s[1]+i+5)=="xxxxx") {
if (string(s[2]+i,s[2]+i+5)=="....x")
if (string(s[3]+i,s[3]+i+5)=="....x")
if (string(s[4]+i,s[4]+i+5)=="xxxxx")
if (string(s[5]+i,s[5]+i+5)=="x....")
if (string(s[6]+i,s[6]+i+5)=="x....")
if (string(s[7]+i,s[7]+i+5)=="xxxxx")
return 2;
}
if (string(s[1]+i,s[1]+i+5)=="xxxxx") {
if (string(s[2]+i,s[2]+i+5)=="....x")
if (string(s[3]+i,s[3]+i+5)=="....x")
if (string(s[4]+i,s[4]+i+5)=="xxxxx")
if (string(s[5]+i,s[5]+i+5)=="....x")
if (string(s[6]+i,s[6]+i+5)=="....x")
if (string(s[7]+i,s[7]+i+5)=="xxxxx")
return 3;
}
if (string(s[1]+i,s[1]+i+5)=="x...x") {
if (string(s[2]+i,s[2]+i+5)=="x...x")
if (string(s[3]+i,s[3]+i+5)=="x...x")
if (string(s[4]+i,s[4]+i+5)=="xxxxx")
if (string(s[5]+i,s[5]+i+5)=="....x")
if (string(s[6]+i,s[6]+i+5)=="....x")
if (string(s[7]+i,s[7]+i+5)=="....x")
return 4;
}
if (string(s[1]+i,s[1]+i+5)=="xxxxx") {
if (string(s[2]+i,s[2]+i+5)=="x....")
if (string(s[3]+i,s[3]+i+5)=="x....")
if (string(s[4]+i,s[4]+i+5)=="xxxxx")
if (string(s[5]+i,s[5]+i+5)=="....x")
if (string(s[6]+i,s[6]+i+5)=="....x")
if (string(s[7]+i,s[7]+i+5)=="xxxxx")
return 5;
}
if (string(s[1]+i,s[1]+i+5)=="xxxxx") {
if (string(s[2]+i,s[2]+i+5)=="x....")
if (string(s[3]+i,s[3]+i+5)=="x....")
if (string(s[4]+i,s[4]+i+5)=="xxxxx")
if (string(s[5]+i,s[5]+i+5)=="x...x")
if (string(s[6]+i,s[6]+i+5)=="x...x")
if (string(s[7]+i,s[7]+i+5)=="xxxxx")
return 6;
}
if (string(s[1]+i,s[1]+i+5)=="xxxxx") {
if (string(s[2]+i,s[2]+i+5)=="....x")
if (string(s[3]+i,s[3]+i+5)=="....x")
if (string(s[4]+i,s[4]+i+5)=="....x")
if (string(s[5]+i,s[5]+i+5)=="....x")
if (string(s[6]+i,s[6]+i+5)=="....x")
if (string(s[7]+i,s[7]+i+5)=="....x")
return 7;
}
if (string(s[1]+i,s[1]+i+5)=="xxxxx") {
if (string(s[2]+i,s[2]+i+5)=="x...x")
if (string(s[3]+i,s[3]+i+5)=="x...x")
if (string(s[4]+i,s[4]+i+5)=="xxxxx")
if (string(s[5]+i,s[5]+i+5)=="x...x")
if (string(s[6]+i,s[6]+i+5)=="x...x")
if (string(s[7]+i,s[7]+i+5)=="xxxxx")
return 8;
}
if (string(s[1]+i,s[1]+i+5)=="xxxxx") {
if (string(s[2]+i,s[2]+i+5)=="x...x")
if (string(s[3]+i,s[3]+i+5)=="x...x")
if (string(s[4]+i,s[4]+i+5)=="xxxxx")
if (string(s[5]+i,s[5]+i+5)=="....x")
if (string(s[6]+i,s[6]+i+5)=="....x")
if (string(s[7]+i,s[7]+i+5)=="xxxxx")
return 9;
}
if (string(s[1]+i,s[1]+i+5)=="xxxxx") {
if (string(s[2]+i,s[2]+i+5)=="x...x")
if (string(s[3]+i,s[3]+i+5)=="x...x")
if (string(s[4]+i,s[4]+i+5)=="x...x")
if (string(s[5]+i,s[5]+i+5)=="x...x")
if (string(s[6]+i,s[6]+i+5)=="x...x")
if (string(s[7]+i,s[7]+i+5)=="xxxxx")
return 0;
}
return -1;
}
void add(int x, int y, string s) {
ans[x][y]=s[0],ans[x][y+1]=s[1],ans[x][y+2]=s[2],ans[x][y+3]=s[3],ans[x][y+4]=s[4];
}
int tot;
void pr(int x) {
++tot;
if (x==1) {
add(1,tot,"....x");
add(2,tot,"....x");
add(3,tot,"....x");
add(4,tot,"....x");
add(5,tot,"....x");
add(6,tot,"....x");
add(7,tot,"....x");
}
if (x==2) {
add(1,tot,"xxxxx");
add(2,tot,"....x");
add(3,tot,"....x");
add(4,tot,"xxxxx");
add(5,tot,"x....");
add(6,tot,"x....");
add(7,tot,"xxxxx");
}
if (x==3) {
add(1,tot,"xxxxx");
add(2,tot,"....x");
add(3,tot,"....x");
add(4,tot,"xxxxx");
add(5,tot,"....x");
add(6,tot,"....x");
add(7,tot,"xxxxx");
}
if (x==4) {
add(1,tot,"x...x");
add(2,tot,"x...x");
add(3,tot,"x...x");
add(4,tot,"xxxxx");
add(5,tot,"....x");
add(6,tot,"....x");
add(7,tot,"....x");
}
if (x==5) {
add(1,tot,"xxxxx");
add(2,tot,"x....");
add(3,tot,"x....");
add(4,tot,"xxxxx");
add(5,tot,"....x");
add(6,tot,"....x");
add(7,tot,"xxxxx");
}
if (x==6) {
add(1,tot,"xxxxx");
add(2,tot,"x....");
add(3,tot,"x....");
add(4,tot,"xxxxx");
add(5,tot,"x...x");
add(6,tot,"x...x");
add(7,tot,"xxxxx");
}
if (x==7) {
add(1,tot,"xxxxx");
add(2,tot,"....x");
add(3,tot,"....x");
add(4,tot,"....x");
add(5,tot,"....x");
add(6,tot,"....x");
add(7,tot,"....x");
}
if (x==8) {
add(1,tot,"xxxxx");
add(2,tot,"x...x");
add(3,tot,"x...x");
add(4,tot,"xxxxx");
add(5,tot,"x...x");
add(6,tot,"x...x");
add(7,tot,"xxxxx");
}
if (x==9) {
add(1,tot,"xxxxx");
add(2,tot,"x...x");
add(3,tot,"x...x");
add(4,tot,"xxxxx");
add(5,tot,"....x");
add(6,tot,"....x");
add(7,tot,"xxxxx");
}
if (x==0) {
add(1,tot,"xxxxx");
add(2,tot,"x...x");
add(3,tot,"x...x");
add(4,tot,"x...x");
add(5,tot,"x...x");
add(6,tot,"x...x");
add(7,tot,"xxxxx");
}
tot += 4;
}
int a[N],cnt;
int main() {
REP(i,1,7) scanf("%s",s[i]+1);
int len = strlen(s[1]+1);
REP(i,1,len) {
a[++cnt] = id(i);
i += 5;
}
ll L = 0, R = 0;
REP(i,1,cnt) {
if (a[i]==-1) {
REP(j,i+1,cnt) R = R*10+a[j];
break;
}
L = L*10+a[i];
}
L += R;
int s[100],cnt=0;
while (L) s[++cnt]=L%10,L/=10;
PER(i,1,cnt) {
pr(s[i]);
if (i!=1) {
++tot;
ans[1][tot]='.';
ans[2][tot]='.';
ans[3][tot]='.';
ans[4][tot]='.';
ans[5][tot]='.';
ans[6][tot]='.';
ans[7][tot]='.';
}
}
REP(i,1,7) puts(ans[i]+1);
}
B. Book Borders
直接二分模拟就能过
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(_i,_a,_n) for(int _i=_a;_i<=_n;++_i)
#define PER(_i,_a,_n) for(int _i=_n;_i>=_a;--_i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(_a) ({REP(_i,1,n) cout<<_a[_i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
const int N = 1e6+50;
int n,a,b,sum[N];
string s[N],t;
int len(int i, int j) {
return sum[j]-sum[i-1]+j-i;
}
int main() {
getline(cin,t);
stringstream ss(t);
while (ss>>t) s[++n]=t;
cin>>a>>b;
REP(i,1,n) sum[i]=sum[i-1]+s[i].size();
REP(x,a,b) {
int t = 1, ans = 0;
while (t<=n) {
int l=t,r=n,pos=-1;
while (l<=r) {
if (len(t,mid)<=x) pos=mid,l=mid+1;
else r=mid-1;
}
if (ans) ans += s[t].size()+1;
else ans += s[t].size();
t = pos+1;
}
printf("%d\n",ans);
}
}
D. Digit Division
求出所有可划分的位置, 如果位置$n$不能划分那么显然无解, 否则为$2^{cnt-1}$
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(_i,_a,_n) for(int _i=_a;_i<=_n;++_i)
#define PER(_i,_a,_n) for(int _i=_n;_i>=_a;--_i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(_a) ({REP(_i,1,n) cout<<_a[_i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
const int N = 1e6+50;
int n,m;
char s[N];
int main() {
scanf("%d%d%s",&n,&m,s+1);
ll t = 0, cnt = 0;
REP(i,1,n) {
t = (t*10+s[i]-'0')%m;
if (!t) ++cnt;
}
if (t) return puts("0"),0;
printf("%d\n",qpow(2,cnt-1));
}
F. Frightful Formula
先考虑$c=0$的情况
对于$t_i$, 考虑从$(2,i)$到$(n,n)$的所有路径, 可以得到贡献为$a^{n-i}b^{n-1}\binom{2n-i-2}{n-i}$.
对于$l_i$, 贡献为$a^{n-1}b^{n-i}\binom{2n-i-2}{n-i}$
然后再考虑$c$的贡献, 相当于是以所有点$(i,j)$为起点,终点为$(n,n)$的路径贡献和
也就是$\sum\limits_{i=2}^n\sum\limits_{j=2}^n a^{n-j}b^{n-i}\binom{2n-i-j}{n-i}$
枚举$i+j$,那么这个式子可以看成卷积,然后用任意模数$NTT$来求.
还有一种方法是设$g_{i,j}=f_{i,j}+\frac{c}{a+b-1}$, 那么$g_{i,j}=ag_{i,j-1}+bg_{i-1,j}$, 这样就可以避免求$c$的贡献
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(_i,_a,_n) for(int _i=_a;_i<=_n;++_i)
#define PER(_i,_a,_n) for(int _i=_n;_i>=_a;--_i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(_a) ({REP(_i,1,n) cout<<_a[_i]<<',';hr;})
using namespace std;
typedef long long ll;
const int P = 1e6+3;
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
const int N = 5e5+10;
int n,a,b,c,l[N],t[N];
int fac[N],ifac[N],pa[N],pb[N];
ll C(int n, int m) {
return (ll)fac[n]*ifac[m]%P*ifac[n-m]%P;
}
int main() {
scanf("%d%d%d%d",&n,&a,&b,&c);
fac[0]=pa[0]=pb[0]=1;
REP(i,1,N-1) {
pa[i]=(ll)pa[i-1]*a%P;
pb[i]=(ll)pb[i-1]*b%P;
fac[i]=(ll)fac[i-1]*i%P;
}
ifac[N-1] = qpow(fac[N-1],P-2);
PER(i,0,N-2) ifac[i]=(ll)ifac[i+1]*(i+1)%P;
int k = (ll)c*qpow(a+b-1,P-2)%P;
REP(i,1,n) scanf("%d",l+i),l[i]=(l[i]+k)%P;
REP(i,1,n) scanf("%d",t+i),t[i]=(t[i]+k)%P;
int ans = 0;
REP(i,2,n) {
ans = (ans+(ll)l[i]*pa[n-1]%P*pb[n-i]%P*C(2*n-i-2,n-i))%P;
ans = (ans+(ll)t[i]*pb[n-1]%P*pa[n-i]%P*C(2*n-i-2,n-i))%P;
}
ans = (ans-k)%P;
if (ans<0) ans+=P;
printf("%d\n",ans);
}
K. Kernel Knights
大意: 给定二分图, 每个点出度为$1$. 求一个核, 满足核内的点之间没边, 核外每个点都与核内有一条方向从内向外的边.
出度为$1$, 那么就保证每个点最多属于一个环, 二分图就保证只有偶环. 所以可以先拓排一下, 最后处理一下偶环即可.
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(_i,_a,_n) for(int _i=_a;_i<=_n;++_i)
#define PER(_i,_a,_n) for(int _i=_n;_i>=_a;--_i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(_a) ({REP(_i,1,n) cout<<_a[_i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
const int N = 1e6+50;
int n, deg[N], nxt[N];
queue<int> q;
int vis[N],ans[N];
int main() {
scanf("%d", &n);
REP(i,1,2*n) scanf("%d",nxt+i),++deg[nxt[i]];
REP(i,1,2*n) if (!deg[i]) q.push(i),vis[i]=1;
while (q.size()) {
int u = q.front(); q.pop();
int v = nxt[u];
if (ans[u]!=2) ans[u] = 1;
if (vis[v]) continue;
if (ans[u]==1) ans[v]=2,vis[v]=1,q.push(v);
if (vis[v]) continue;
if (!--deg[v]) vis[v]=1,q.push(v);
}
REP(i,1,2*n) if (!vis[i]) {
int now = i, cur = 0;
while (!vis[now]) {
cur ^= 1;
vis[now] = 1;
if (cur) ans[now] = 1;
now = nxt[now];
}
}
REP(i,1,2*n) if (ans[i]==1) printf("%d ",i);hr;
}