https://atcoder.jp/contests/abc202/tasks/abc202_d

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思路：后往前考虑当前位取不取1，然后取得话我要知道这时候放了1，然后这个用了多少的部分用dp求。
dp[i][j]：到前i个且有j个1的总数。考虑第i位是1还是0

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=70;
typedef long long LL;
inline LL read(){LL x=0,f=1;char ch=getchar();	while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;}
LL dp[maxn][maxn];
void pre(){
    for(LL i=0;i<maxn;i++){
        for(LL j=0;j<=i;j++){
            if(!j) dp[i][j]=1;
            else dp[i][j]=dp[i-1][j-1]+dp[i-1][j];
        }
    }
}
int main(void){
   cin.tie(0);std::ios::sync_with_stdio(false);
   pre();
   LL a,b,k;cin>>a>>b>>k;
   LL num=b;
   for(LL i=a+b;i>=1;i--){
       if(num>0&&dp[i-1][num]<k){
          cout<<"b";
          k-=dp[i-1][num];
          num--;
       }
       else cout<<"a";
   }
   cout<<"\n";
   return 0;
}