https://codeforces.com/contest/44/problem/E
高级又好像很低级的诈骗
首先不难得到 \(a\times k>|s|\texttt{ or }b\times k<|s|\) 无解。
对于每一组考虑先填上 \(a\) 个字符，对于多出来的依次枚举把能填上（设当前剩 \(p\) 个，则能填上的个数为 \(\min(b-a,p)\)）的填上即可。
# include <bits/stdc++.h>
using namespace std;
int main () {
ios :: sync_with_stdio (false);
cin .tie (0);
cout .tie (0);
int k, a, b;
string s;
cin >> k >> a >> b >> s;
int n = s .size ();
if (n < k * a || n > k * b) {
cout << "No solution";
return 0;
}
int p = n - k * a, u = 0;
for (int i = 1; i <= k; ++ i) {
for (int j = 1; j <= a; ++ j, ++ u) {
cout << s [u];
}
for (int j = 1; j <= min (p, b - a); ++ j, ++ u) {
cout << s [u];
}
p -= min (p, b - a);
cout << '\n';
}
return 0;
}