【POJ1011】Sticks

【题目概括】

现在有\(n\)个长度不超过\(50\)的木棍，请你把这些小木棍拼成若干根长度相同的木棍。

请你最小化拼成后的长度。

【思路要点】

【代码】

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>

#define FI first
#define SE second
#define REP(i, s, t) for (int i = s; i <= t; i++)
#define PER(i, s, t) for (int i = s; i >= t; i--)
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef long double ld;

template <class T> void chkmax(T& x, T y) { x = max(x, y); }
template <class T> void chkmin(T& x, T y) { x = min(x, y); }

namespace input {
  template <class T>
  void read(T& x) {
    x = 0; char ch = 0; int f = 1;
    for (; !isdigit(ch); ch = getchar()) if (ch == '-') f = -1;
    for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    x *= f;
  }
  void re(int& x) { read(x); }
  void re(ll& x) { read(x); }
  void re(ull& x) { read(x); }
  void re(char& x) { x = getchar(); }
  void re(string& x) { cin >> x; }
}
using namespace input;

namespace output {
  template <class T>
  void write(T x) {
    if (!x) { putchar('0'); return; }
    if (x < 0) putchar('-'), x = -x;
    static int top, stk[25]; top = 0;
    while (x) stk[++top] = x % 10, x /= 10;
    while (top) putchar(stk[top--] + 48);
  }
  void pr(int x) { write(x); }
  void pr(ll x) { write(x); }
  void pr(ull x) { write(x); }
  void pr(char x) { putchar(x); }
  void pr(string x) { cout << x; }
  void pp() { putchar(' '); }
  void ps() { puts(""); }
}
using namespace output;

const int N = 105;

int n, mx, mi, sum, need;
int a[N], buk[N];
bool flg = 0;

void dfs(int rest, int nowLen, int lastChose) {
  if (!rest) {
    pr(need), pr('\n');
    flg = 1;
    return;
  }
  if (nowLen == need) {
    dfs(rest - 1, 0, mx);
    return;
  }
  for (int i = lastChose; i >= mi; i--) {
    if (flg)
      return;
    if (buk[i] && nowLen + i <= need) {
      buk[i]--;
      dfs(rest, nowLen + i, i);
      buk[i]++;
      if (!nowLen || nowLen + i == need)
        break;
    }
  }
}

void solve() {
  memset(buk, 0, sizeof buk);
  sum = mx = 0; flg = 0; mi = inf;
  for (int i = 1, x; i <= n; i++) {
    re(x), sum += x, buk[x]++, chkmax(mx, x), chkmin(mi, x);
  }
  for (int i = mx; i <= sum / 2; i++)
    if (sum % i == 0) {
      need = i;
      dfs(sum / i, 0, mx);
      if (flg)
        break;
    }
  if (!flg)
    pr(sum), pr('\n');
}

int main() {
  while (1) {
    re(n);
    if (!n)
      return 0;
    solve();
  }
  return 0;
}