gym101480
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; }
