POJ1011 Sticks

来源：互联网发布：时间同步软件编辑：程序博客网时间：2024/05/17 23:31
#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <vector>#include <list>#include <deque>#include <queue>#include <cctype>#include <map>#include <set>#include <bitset>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iomanip>#include <cstdlib>#include <ctime>#include <cassert>#include <limits>#include <fstream>using namespace std;#define mem(A, X) memset(A, X, sizeof A)#define pb(x) push_back(x)#define mp(x,y) make_pair((x),(y))#define all(x) x.begin(), x.end()#define sz(x) (int)((x).size())#define rep(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))#define Rep(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))#define min3(a,b,c) min(a,min(b,c))#define max3(a,b,c) max(a,max(b,c))#define dbg(a) cout << a << endl;#define fi first#define se secondtypedef long long int64;int gcd(const int64 &a, const int64 &b) {return b == 0 ? a : gcd(b, a % b);}int64 int64pow(int64 a, int64 b){if(b == 0) return 1;int64 t = int64pow(a, b / 2);if(b % 2) return t * t * a;return t * t;}const int inf = 1 << 30;const double eps = 1e-8;const double pi = acos(-1.0);const int MAX_N = 1005;int arr[MAX_N], sum, n, ans, MAX;bool visit[MAX_N];bool comp(const int &a, const int &b){    return a > b;}bool dfs(int *arr, bool *visit, int len, int ans, int s, int num){    if (num == n)        return true;    int sample = -1;    rep(i, s, n) {        if (visit[i] || arr[i] == sample)//剪枝3:等长的木棒只搜索一次            continue;        visit[i] = true;        if (len + arr[i] < ans) {            if (dfs(arr, visit, len + arr[i], ans, i, num + 1))                return true;            else                sample = arr[i];        }        else if (len + arr[i] == ans) {            if (dfs(arr, visit, 0, ans, 0, num + 1))                return true;            else                sample = arr[i];        }        visit[i] = false;        if (len == 0)//剪枝4:构建新棒时,对于新棒的第一根棒子，在搜索完所有棒子后都无法组合            break;  //则说明该棒子无法在当前组合方式下组合,不用往下搜索(往下搜索会令该棒子被舍弃),直接返回上一层    }    return false;}void solve(){    mem(visit, 0);    sum = ans = 0;    rep(i, 0, n) {        scanf("%d", &arr[i]);        sum += arr[i];    }    sort(arr, arr + n, comp);    MAX = arr[0];    Rep(i, 1, sum) { //剪枝1:答案必然在[1, sum]内        if ((sum % i == 0) && dfs(arr, visit, 0, i, 0, 0)) {//剪枝2: 答案必须是sum的约数            ans = i;            break;        }    }    printf("%d\n", ans);}int main(){    while (cin >> n && n) {        solve();    }}
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