预处理: dp[i][j][k]表示i行诗,第j层,最大字母是k有多少个。
然后直接dfs找就行了。
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define LL bign
#define pii pair<int,int>
#define SZ(x) (int)x.size()
#define all(x) x.begin(),x.end()
using namespace std;
const int MAXN=150;
struct bign
{
int len, s[MAXN];
bign ()
{
memset(s, 0, sizeof(s));
len = 1;
}
bign (int num) { *this = num; }
bign (const char *num) { *this = num; }
bign operator = (const int num)
{
char s[MAXN];
sprintf(s, "%d", num);
*this = s;
return *this;
}
bign operator = (const char *num)
{
for(int i = 0; num[i] == '0'; num++) ; //去前导0
len = strlen(num);
for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
return *this;
}
bign operator + (const bign &b) const //+
{
bign c;
c.len = 0;
for(int i = 0, g = 0; g || i < max(len, b.len); i++)
{
int x = g;
if(i < len) x += s[i];
if(i < b.len) x += b.s[i];
c.s[c.len++] = x % 10;
g = x / 10;
}
return c;
}
bign operator += (const bign &b)
{
*this = *this + b;
return *this;
}
void clean()
{
while(len > 1 && !s[len-1]) len--;
}
bign operator * (const bign &b) //*
{
bign c;
c.len = len + b.len;
for(int i = 0; i < len; i++)
{
for(int j = 0; j < b.len; j++)
{
c.s[i+j] += s[i] * b.s[j];
}
}
for(int i = 0; i < c.len; i++)
{
c.s[i+1] += c.s[i]/10;
c.s[i] %= 10;
}
c.clean();
return c;
}
bign operator *= (const bign &b)
{
*this = *this * b;
return *this;
}
bign operator - (const bign &b)
{
bign c;
c.len = 0;
for(int i = 0, g = 0; i < len; i++)
{
int x = s[i] - g;
if(i < b.len) x -= b.s[i];
if(x >= 0) g = 0;
else
{
g = 1;
x += 10;
}
c.s[c.len++] = x;
}
c.clean();
return c;
}
bign operator -= (const bign &b)
{
*this = *this - b;
return *this;
}
bign operator / (const bign &b)
{
bign c, f = 0;
for(int i = len-1; i >= 0; i--)
{
f = f*10;
f.s[0] = s[i];
while(f >= b)
{
f -= b;
c.s[i]++;
}
}
c.len = len;
c.clean();
return c;
}
bign operator /= (const bign &b)
{
*this = *this / b;
return *this;
}
bign operator % (const bign &b)
{
bign r = *this / b;
r = *this - r*b;
return r;
}
bign operator %= (const bign &b)
{
*this = *this % b;
return *this;
}
bool operator < (const bign &b)
{
if(len != b.len) return len < b.len;
for(int i = len-1; i >= 0; i--)
{
if(s[i] != b.s[i]) return s[i] < b.s[i];
}
return false;
}
bool operator > (const bign &b)
{
if(len != b.len) return len > b.len;
for(int i = len-1; i >= 0; i--)
{
if(s[i] != b.s[i]) return s[i] > b.s[i];
}
return false;
}
bool operator == (const bign &b)
{
return !(*this > b) && !(*this < b);
}
bool operator != (const bign &b)
{
return !(*this == b);
}
bool operator <= (const bign &b)
{
return *this < b || *this == b;
}
bool operator >= (const bign &b)
{
return *this > b || *this == b;
}
string str() const
{
string res = "";
for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;
return res;
}
};
istream& operator >> (istream &in, bign &x)
{
string s;
in >> s;
x = s.c_str();
return in;
}
ostream& operator << (ostream &out, const bign &x)
{
if (x.str()=="") out<<0;
else out << x.str();
return out;
}
const int N = 5e5 + 3;
int t,n,now,cnt;
LL k;
char a[33];
LL dp[30][30][30];
bign zero=0;
void dfs(int now,char x,int pos,char M='A'){
if(dp[pos][now][M-'A']!=zero){return ;}
dp[pos][now][M-'A']=0;
if(now==pos){dp[pos][now][M-'A']=1;return;}
for(char i='A';i<=M+1;i++){
dfs(now+1,i,pos,max(M,i));
dp[pos][now][M-'A']+=dp[pos][now+1][max(i,M)-'A'];
}
}
void get(int now,char x,char M='A'){
a[cnt++]=x;
if(now==n)return ;
LL pre=0;
for(char i='A';i<=M+1;i++){
if(pre+dp[n][now+1][max(M,i)-'A']<k){
pre+=dp[n][now+1][max(M,i)-'A'];
}else{
k-=pre;
get(now+1,i,max(M,i));
break;
}
}
}
int main() {
ios::sync_with_stdio(false);
for(int i=1;i<=26;i++){
for(int j=0;j<=26;j++){
for(int k=0;k<26;k++)dp[i][j][k]=0;
}
dfs(1,'A',i);
}
for(cin>>t;t;t--){
cin>>n>>k;
cnt=0;
get(1,'A');
cout<<"Case #"<<++now<<": ";
for(int i=0;i<n;i++)cout<<a[i];
cout<<'\n';
}
return 0;
}