Rhyme scheme

最新推荐文章于 2025-06-19 10:45:14 发布

转载最新推荐文章于 2025-06-19 10:45:14 发布 · 985 阅读

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原文链接：http://www.cnblogs.com/czy-power/p/11525174.html

31.43%
1000ms
65536K

A rhyme scheme is the pattern of rhymes at the end of each line of a poem or song. It is usually referred to by using letters to indicate which lines rhyme; lines designated with the same letter all rhyme with each other.

e.g., the following "poem'' of

1 ——

2 —— Fix one line B

3 —— Should be fine B

4 ——

This essentially means that line

The number of different possible rhyme schemes for an $B_3 = 5B3=5, it means there are five rhyme schemes for a three-line poem: AAA, AAB, ABA, ABB, and ABC.$

The question is to output the

InputFile

The first line of the input gives the number of test cases, $\leq T \leq 100001≤T≤10000). TT test cases follow.$

Each test case contains a line with two integers

$\leq n \leq 26, 1 \leq k \leq B_n1≤n≤26,1≤k≤Bn (B_nBn is the nn-th of Bell numbers)$

OutputFile

For each test case, output one line containing Case #x: y, where

样例输入复制

样例输出复制

Case #1: A
Case #2: AA
Case #3: AAA
Case #4: AAB
Case #5: ABA
Case #6: ABB
Case #7: ABC

编辑代码

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int mod = 1000000007;
const int maxn = 305;

int T, n;
__int128 k;
__int128 dp[36][36];
int res[maxn];

inline __int128 read() {
    __int128 x = 0, f = 1;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -1;
    for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
    return x * f;
}


inline __int128 dfs1(int id, int limit) {
    if (id == n + 1)return 1;
    if (~dp[id][limit])
        return dp[id][limit];
    __int128 tot = 0;
    for (register int i = 0; i <= limit; ++i) {
        tot += dfs1(id + 1, max(limit, i + 1));
    }
    return dp[id][limit] = tot;
}

inline void solve(int id, int limit, __int128 leave) {
    if (id == n + 1) {
        for (register int i = 1; i <= n; ++i) {
            printf("%c", res[i] + 'A');
        }
        printf("\n");
        return;
    }
    for (register int i = 0; i <= limit; ++i) {
        __int128 cur = max(limit, i + 1);
        if (dp[id + 1][cur] >= leave) {
            res[id] = i;
            solve(id + 1, cur, leave);
            break;
        } else {
            leave -= dp[id + 1][cur];
        }
    }
}

int Case;

int main() {
#ifndef ONLINE_JUDGE
    //freopen("1.txt", "r", stdin);
#endif
    scanf("%d", &T);
    while (T--) {
        scanf("%d%lld", &n,&k);
        //k = read();
        for (register int i = 0; i <= 30; ++i) {
            for (register int j = 0; j <= 30; ++j) {
                dp[i][j] = -1;
            }
        }
        for (register int i = 0; i <= 30; ++i)dp[n + 1][i] = 1;
        dfs1(1, 0);
        printf("Case #%d: ", ++Case);
        solve(1, 0, k);
    }
    return 0;
}

转载于:https://www.cnblogs.com/czy-power/p/11525174.html