1. Abridged Problem Statement  
Given a non-negative integer Q (0 ≤ Q ≤ 10^8), find the smallest positive integer N such that the number of trailing zeros in N! (in base 10) is exactly Q. If no such N exists, print “No solution.”.

2. Detailed Editorial  
Overview  
• Trailing zeros in N! arise from factors of 10, i.e. pairs of 2×5. Since factorials have many more factors of 2 than 5, the count of trailing zeros is governed by the number of times 5 divides into the product 1×2×…×N.  
• The standard formula for the number of trailing zeros in N! is:  
  Z(N) = ⌊N/5⌋ + ⌊N/5^2⌋ + ⌊N/5^3⌋ + …  
  where terms stop when 5^k > N.  

Monotonicity and Search  
• Z(N) is non-decreasing as N increases. It jumps by 1 at values of N that are multiples of 5 (more at multiples of higher powers of 5).  
• To find the minimal N with Z(N) = Q (if it exists), we can binary-search on N over a sufficiently large range.  

Key Steps  
1. Define eval(x) = Z(x).  
2. Binary search low = 1, high = 2×10^12 (safe upper bound for Q up to 10^8).  
   – At each step, compute mid = (low+high)//2.  
   – If eval(mid) ≥ Q, record mid as a candidate and move high = mid−1.  
   – Otherwise, move low = mid+1.  
3. After search, check if eval(candidate) == Q.  
   – If yes, print candidate.  
   – Otherwise, print “No solution.”.  

Complexity  
• Computing eval(x) takes O(log_5 x) steps.  
• Binary search does O(log high) iterations.  
Overall O(log high · log_5 high), fast enough under 0.25 s.

3. Provided C++ Solution with Line-by-Line Comments  
#include <bits/stdc++.h>  
using namespace std;  
// Read q, compute minimal N so that N! has exactly q trailing zeros.

template<class T, class T2>
inline int chkmax(T& x, const T2& y) {
    return x < y ? x = y, 1 : 0;
}
template<class T, class T2>
inline int chkmin(T& x, const T2& y) {
    return x > y ? x = y, 1 : 0;
}

int q;

// Read input Q
void read() {
    cin >> q;
}

// Evaluate the number of trailing zeros in x!
int64_t eval(int64_t x) {
    int64_t ret = 0;   // result: count of zeros
    int64_t c = 0;     // cumulative sum of floor(x/5^i) processed so far
    int64_t cc = 1;    // current exponent index: how many powers of 5 fit
    int64_t l = 5;     // current power of 5

    // First find the highest power of 5 <= x, and cc = number of terms
    while (l <= x) {
        cc++;
        l *= 5; 
    }
    // Now l is first power of 5 > x, cc is one too large
    // Decrease cc, divide l by 5, and accumulate floors
    while (l > 1) {
        // floor(x/l) gives how many multiples of current power of 5
        ret += cc * ( (x / l) - c );
        c += (x / l) - c;  // update cumulative count
        cc--;              // next lower power
        l /= 5;            // step down power of 5
    }
    return ret;
}

void solve() {
    int64_t low = 1, high = (int64_t)2e12, mid, ret = -1;
    // Binary search for minimal x with eval(x) >= q
    while (low <= high) {
        mid = (low + high) >> 1;
        if (eval(mid) >= q) {
            ret = mid;       // candidate found
            high = mid - 1;  // try smaller
        } else {
            low = mid + 1;   // need larger x
        }
    }
    // Check exact match
    if (ret == -1 || eval(ret) != q) {
        cout << "No solution." << '\n';
    } else {
        cout << ret << '\n';
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    read();
    solve();
    return 0;
}

4. Python Solution with Detailed Comments  
```python
import sys

def eval_zeros(x):
    """Return the number of trailing zeros in x!."""
    total = 0
    power = 5
    # Keep dividing x by powers of 5
    while power <= x:
        total += x // power
        power *= 5
    return total

def find_min_n(q):
    """Binary-search minimal n so that eval_zeros(n) >= q."""
    low, high = 1, 2 * 10**12
    answer = -1
    while low <= high:
        mid = (low + high) // 2
        if eval_zeros(mid) >= q:
            answer = mid
            high = mid - 1  # search lower half
        else:
            low = mid + 1   # search upper half
    return answer

def main():
    data = sys.stdin.read().strip()
    if not data:
        return
    q = int(data)
    n = find_min_n(q)
    # Verify exact match
    if n == -1 or eval_zeros(n) != q:
        print("No solution.")
    else:
        print(n)

if __name__ == "__main__":
    main()
```

5. Compressed Editorial  
Count trailing zeros via Z(N)=∑⌊N/5^i⌋. Because Z(N) is non-decreasing, binary-search N in [1,2×10^12] to find the smallest N with Z(N)=Q. If none matches exactly, report no solution.