字典顺序排在第k位(只有a,b)

Among the strings of length A+B containing A occurrences of a and B occurrences of b, find the string that comes K-th in the lexicographical order.

Constraints
1≤A,B≤30
1≤K≤S, where S is the number of strings of length A+B containing A occurrences of a and B occurrences of b.
All values in input are integers.

输入

Input is given from Standard Input in the following format:

A B K

输出

Print the answer.

样例输入 Copy

【样例1】
2 2 4
【样例2】
30 30 118264581564861424

样例输出 Copy

【样例1】
baab
【样例2】
bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

提示

样例1解释
Here are the strings containing two as and two bs in the lexicographical order: aabb, abab, abba, baab, baba, and bbaa. The fourth string, baab, should be printed.
样例2解释
K may not fit into a 32-bit integer type.

 

 

题意：这个题就是构造一个长度为A+B的字符串，其中包含A个a和B个b，并且字典序为第k位。

这个题就是一个组合数的问题：

构造一个全是a的字符串，然后从大到小枚举每一个字符，放上b，

如果说i是放第i个字符，z是第z位。

c=C[z][b-i+1];
d=C[z+1][b-i+1];

这个就假如b放在第i位后面还有k个字符,其中有z个b字符，然后他的字典数位c[z][k]

就是这样判断的

#include<iostream>
#include<algorithm>
#include<cstring>
typedef long long ll;
using namespace std;
const int maxn=1e3+100;
int b[maxn];
ll C[maxn][maxn];
void get_C()
{
    C[0][0] = 1;
    for(int i=1;i<=100;i++)
    {
        C[i][0] = 1;
        for(int j=1;j<=i;j++)
            C[i][j] = C[i-1][j]+C[i-1][j-1];
    }
}
int main(){
    get_C();
    ll a,b,k;
    cin>>a>>b>>k;
    k--;
    ll n=a+b; 
    vector<char> v(n,'a');
    int p=a;
    ll z=n;
    for(int i=1;i<=b;i++){
        ll c,d;
        for(;z>=1;z--){
            c=C[z][b-i+1];
            d=C[z+1][b-i+1];
            if(c <= k && d > k){
                break;
            }
        }
        k-=c;
        v[z]='b';
    }
    reverse(v.begin(),v.end());
    for(int i = 0; i < n; i++) cout << v[i];
    cout << endl;
}