枚举每一位, 用dp去算方案数。
#include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; const int N = 20 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;} LL F[18]; LL dp[16][18]; int c[16]; int ans[17]; int k, t; LL getDp(int n, int *c) { if(n == 0) return 1; memset(dp, 0, sizeof(dp)); for(int i = 0; i <= c[0]; i++) dp[0][i] = 1; for(int i = 0; i < 15; i++) { for(int j = 0; j <= n; j++) { for(int k = 0; k <= c[i + 1]; k++) { if(j + k > n) continue; dp[i + 1][j + k] += F[j + k] / F[j] / F[k] * dp[i][j]; } } } return dp[15][n]; } char print(int x) { if(x <= 9) return '0' + x; else return 'a' + (x - 10); } int main() { for(int i = F[0] = 1; i <= 17; i++) F[i] = F[i - 1] * i; scanf("%d%d", &k, &t); for(int i = 0; i <= 15; i++) c[i] = t; for(int len = 1; len <= 17; len++) { LL tot = 0; for(int j = 1; j <= 15; j++) { if(!c[j]) continue; c[j]--; int way = getDp(len - 1, c); if(way < k) k -= way, c[j]++, tot += way; else { ans[1] = j; break; } } if(!ans[1]) continue; for(int i = 2; i <= len; i++) { for(int j = 0; j <= 15; j++) { if(!c[j]) continue; c[j]--; int way = getDp(len - i, c); if(way < k) k -= way, c[j]++; else { ans[i] = j; break; } } } for(int i = 1; i <= len; i++) printf("%c", print(ans[i])); puts(""); return 0; } return 0; } /* */