NAIPC 2018
E. Prefix Free Code
大意: 给定$n$个串, 保证任意一个串都不是另一个串的前缀, 从中选出$k$个串可以拼成$\binom{n}{k}k!$种串. 给定其中一个串, 求这个串的排名.
先用字典树处理一下, 从而转化为给定一个$n$元素中取$k$元素的排列, 求排名.
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#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 = 1e9+7; const int N = 1e6+10; ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} int n,k,ch[N][26],T,val[N],a[N],tot,cnt,len; char s[N]; void add(int &o, char *s) { if (!o) o = ++tot; if (*s) add(ch[o][*s-'a'],s+1); else val[o] = 1; } void dfs(int o) { if (!o) return; if (val[o]) return val[o] = ++cnt,void(); REP(i,0,25) dfs(ch[o][i]); } void find(int o, char *s) { if (val[o]) a[++*a]=val[o]; else ++len,find(ch[o][*s-'a'],s+1); } int c[N]; void add(int x, int v) { for (; x<=n; x+=x&-x) c[x]+=v; } int qry(int x) { int ret = 0; for (; x; x^=x&-x) ret+=c[x]; return ret; } int main() { scanf("%d%d",&n,&k); REP(i,1,n) scanf("%s",s),add(T,s); dfs(T); scanf("%s",s+1); int m = strlen(s+1); REP(i,1,m) len=0,find(T,s+i),i+=len-1; int tot = 1; REP(i,n-k+1,n) tot = (ll)tot*i%P; REP(i,1,n) c[i] = i&-i; int ans = 1; REP(i,1,k) { tot = (ll)tot*inv(n-i+1)%P; ans = (ans+(ll)qry(a[i]-1)*tot)%P; add(a[i],-1); } printf("%d\n",ans); }
H. Recovery
大意: 给定每行每列和的奇偶性, 要求恢复一个$n\times m$的$01$矩阵, 若有多种方案输出$1$个数最多的, 还有多种的话输出字典序最小的.
先初始化全$1$, 那么就转化为要改变某些行某些列的奇偶性, 如果行列需要改动的个数之差为奇数则无解, 偶数时贪心构造一下.
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#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 = 55; int n,m; vector<int> va,vb; char ans[N][N],a[N],b[N]; int main() { scanf("%s%s",a+1,b+1); n = strlen(a+1); m = strlen(b+1); REP(i,1,n) if (m%2==0&&a[i]=='1'||m%2==1&&a[i]=='0') va.pb(i); REP(i,1,m) if (n%2==0&&b[i]=='1'||n%2==1&&b[i]=='0') vb.pb(i); int sza = va.size(), szb = vb.size(); if ((sza-szb)&1) return puts("-1"),0; REP(i,1,n) REP(j,1,m) ans[i][j]='1'; if (szb>=sza) { REP(i,0,szb-sza-1) ans[1][vb[i]] = '0'; int now = 0; REP(i,szb-sza,szb-1) ans[va[now++]][vb[i]] = '0'; } else { REP(i,0,sza-szb-1) ans[va[i]][1] = '0'; int now = 0; REP(i,sza-szb,sza-1) ans[va[i]][vb[now++]] = '0'; } REP(i,1,n) puts(ans[i]+1); }