Solution

如果单看这个递推式，可以很简单的发现除开第一项和第二项，从第三项开始都是自己加上前一项的结果（字符串拼接）作为自己的答案。因为n比较小，我选择模拟去算。那么就要设计到如何递推了，单纯变量肯定不行，因为涉及字母与次数的映射，选择用去循环模拟。

Code

#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;

inline int read() {
    int s = 0, w = 1; char ch = getchar();
    while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }
    while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
    return s * w;
}

map<char, ll> mp1, mp2, tmp;
string a, b;
int n;

int main() {
    js;
    cin >> a >> b >> n;
    for (auto it : a)
        ++mp1[it];
    if (n == 1) { //注意1，2特判
        for (auto it : mp1)
            cout << it.first << ": " << it.second << endl;
        return 0;
    }
    for (auto it : b)
        ++mp2[it];
    if (n == 2) {
        for (auto it : mp2)
            cout << it.first << ": " << it.second << endl;
        return 0;
    }
    n -= 2;
    while (n > 0) {
        tmp = mp2;
        --n;
        for (auto it : mp1)
            mp2[it.first] += mp1[it.first];
        mp1 = tmp;
    }
    for (auto it : mp2)
        cout << it.first << ": " << it.second << endl;
    return 0;
}