Comet OJ 【noip模拟Day2】T2
一道类似康拓展开的数学题，结合了字符串的考点
这是题解的下载链接click here
难点在于如何判断字符串是由哪些单词组成的，正解是字典树，也可以hash，用map能拿90分，手写hash表可以AC
这里给出map代码和手写hash代码
STL map
#include<bits/stdc++.h>
using namespace std;
#define go(i,a,b) for(int i=a;i<=b;++i)
#define com(i,a,b) for(int i=a;i>=b;--i)
#define mem(a,b) memset(a,b,sizeof(a))
#define int long long
#define lowbit(x) (x&-x)
#pragma GCC optimize(2)
const int inf=0x3f3f3f3f,mod=1e9+7,N=1e6+10;
int n,m,len[N],jc[N],inv[N],c[N];
bool vis[N];
string s[N],b;
map<string,int>id;
inline int mul(const int &x,const int &y){
return x*y>=mod?x*y%mod:x*y;
}
inline int add(const int &x,const int &y){
return x+y>=mod?x+y-mod:x+y;
}
inline void init(){
cin>>n>>m;
jc[0]=1;
go(i,1,n) jc[i]=mul(jc[i-1],i);
inv[0]=inv[1]=1;
go(i,2,n) inv[i]=mul((mod-mod/i),inv[mod%i]);
go(i,2,n) inv[i]=mul(inv[i],inv[i-1]);
go(i,1,n) cin>>s[i];
sort(s+1,s+n+1);
go(i,1,n){
id[s[i]]=i;//标号必须在排序之后标号，否则会全WA
len[i]=s[i].length();
}
cin>>b;
}
inline int C(int n,int m){
return jc[n]*inv[m]%mod*inv[n-m]%mod;
}
inline void add(int x){
for(;x<=n;x+=lowbit(x)) ++c[x];
}
inline int ask(int x){
int ans=0;
for(;x;x-=lowbit(x)) ans+=c[x];
return ans;
}
signed main(){
init();
int ans=0;
string c;
int i=0;
for(int pos=0;pos<b.length();++pos){
int cnt=0;
c+=b[pos];
if(id.count(c)){
++i;
int j=id[c];
string tmp;
swap(c,tmp);
add(j);
cnt=j-ask(j-1)-1;
ans=add(ans,mul(mul(cnt,C(n-i,m-i)),jc[m-i]));
}
}
printf("%lld",(ans+1)%mod);
return 0;
}
手写hash
#include<bits/stdc++.h>
using namespace std;
#define go(i,a,b) for(int i=a;i<=b;++i)
#define com(i,a,b) for(int i=a;i>=b;--i)
#define mem(a,b) memset(a,b,sizeof(a))
#define lowbit(x) (x&-x)
#define int long long
const int inf=0x3f3f3f3f,mod=1e9+7,N=1e6+10,P=1e7+13;
typedef long long ll;
int n,m,len[N],jc[N],inv[N],c[N],has[N],head[N*10],cn=0;
string s[N],b;
struct edge{
int nxt,p;
}e[N];
void insert(int x,int p){
e[cn]=(edge){head[x],p};
head[x]=cn++;
}
int query(int x,int pos){
for(int i=head[x];i+1;i=e[i].nxt){
int p=e[i].p;
bool ok=1;
for(int j=len[p]-1,k=pos;j>=0;--j,--k){
if(s[p][j]!=b[k]){
ok=0;
break;
}
}
if(ok) return p;
}
return 0;
}
inline int add(const int &x,const int &y){
return x+y>=mod?x+y-mod:x+y;
}
inline int mul(const int &x,const int &y){
return 1ll*x*y>=mod?1ll*x*y%mod:x*y;
}
inline void init(){
cin>>n>>m;
jc[0]=1;
go(i,1,n) jc[i]=mul(jc[i-1],i);
inv[0]=inv[1]=1;
go(i,2,n) inv[i]=mul((mod-mod/i),inv[mod%i]);
go(i,2,n) inv[i]=mul(inv[i],inv[i-1]);
go(i,1,n) cin>>s[i];
sort(s+1,s+n+1);
go(i,1,n){
len[i]=s[i].length();
for(int j=0;j<len[i];++j){
has[i]=((ll)has[i]*131+(ll)(s[i][j]))%P;
}
insert(has[i],i);
}
cin>>b;
}
inline int C(int n,int m){
return mul(mul(jc[n],inv[m]),inv[n-m]);
}
inline void add(int x){
for(;x<=n;x+=lowbit(x)) ++c[x];
}
inline int ask(int x){
int ans=0;
for(;x;x-=lowbit(x)) ans+=c[x];
return ans;
}
signed main(){
//freopen("input.txt","r",stdin);
mem(head,-1);
init();
int c=0;
int ans=0,i=0,j,l=b.length(),cnt;
for(int pos=0;pos<l;++pos){
cnt=0;
c=((ll)c*131+(ll)b[pos])%P;
if((j=query(c,pos))){
++i;
c=0;
add(j);
cnt=j-ask(j-1)-1;
ans=add(ans,mul(mul(cnt,C(n-i,m-i)),jc[m-i]));
}
}
printf("%lld",(ans+1)%mod);
return 0;
}
用字典树更快，这里给出一位同学的代码
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ll;
const int maxn=1e6+5;
const int mod=1e9+7;
int n,k,cnt;
int siz[maxn*26],ch[maxn][30],fac[maxn];
int ok[maxn*26];
char s[maxn],c[maxn];
void ins(char s[]){
ll len=strlen(s+1),rt=1;
for(ll i=1;i<=len;i++){
ll x=s[i]-'a'+1;
++siz[rt];
if(!ch[rt][x]) ch[rt][x]=++cnt;
rt=ch[rt][x];
}
ok[rt]=1,++siz[rt];
}
void solve(){
ll len=strlen(c+1),rt=1,ans=0;
for(ll i=1;i<=len;i++){
ll x=c[i]-'a'+1,ord=0;
--siz[rt];
for(ll j=1;j<x;j++) ord=(ord+siz[ch[rt][j]])%mod;
ans=(ans+ord*fac[k]%mod)%mod;
rt=ch[rt][x];
if(ok[rt]){
--k;
--siz[rt];
rt=1;
}
}
printf("%d",(ans+1)%mod);
}
int main(){
//freopen("input.txt","r",stdin);
scanf("%d%d",&n,&k);
fac[1]=cnt=1;
for(ll i=2,j=n-k+1;j<=n-1;i++,j++){
fac[i]=1ll*fac[i-1]*j%mod;
}
for(ll i=1;i<=n;i++){
scanf("%s",s+1);
ins(s);
}
scanf("%s",c+1);
solve();
return 0;
}
