SRM 622 D2L3: Subsets, math, backtrack

题目：http://community.topcoder.com/stat?c=problem_statement&pm=10554&rd=15855

符合条件的集中非1的元素个数是非常少的，能够用回溯加剪枝。实际执行速度非常快。

代码：

#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <iostream>
#include <sstream>
#include <iomanip>

#include <bitset>
#include <string>
#include <vector>
#include <stack>
#include <deque>
#include <queue>
#include <set>
#include <map>

#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <cstring>
#include <ctime>
#include <climits>
using namespace std;

#define CHECKTIME() printf("%.2lf\n", (double)clock() / CLOCKS_PER_SEC)
typedef pair<int, int> pii;
typedef long long llong;
typedef pair<llong, llong> pll;
#define mkp make_pair
#define FOREACH(it, X) for(__typeof((X).begin()) it = (X).begin(); it != (X).end(); ++it)

/*************** Program Begin **********************/

class Subsets {
public:
vector <int> numbers;
int ones_cnt;
int res;
int nextdiff[1005];
void backtrack(int sum, int prod, int pos)
{
// add
int cur = numbers[pos];
int next_sum = sum + cur;
int next_prod = prod * cur;
if (next_sum + ones_cnt > next_prod) {
res += next_sum + ones_cnt - next_prod;
if (pos + 1 < numbers.size()) {
backtrack(next_sum, next_prod, pos + 1);
}
}

// not add
if (nextdiff[pos] < numbers.size()) {
backtrack(sum, prod, nextdiff[pos]);
}
}

int findSubset(vector <int> numbers) {
sort(numbers.begin(), numbers.end());
this->numbers = numbers;
int n = numbers.size();

for (int i = 0; i < n; i++) {
nextdiff[i] = n;
for (int j = i + 1; j < n; j++) {
if (numbers[i] != numbers[j]) {
nextdiff[i] = j;
break;
}
}
}

ones_cnt = count(numbers.begin(), numbers.end(), 1);
res = max(ones_cnt - 1, 0);
if (ones_cnt < n) {
backtrack(0, 1, ones_cnt);
}

return res;
}

};

/************** Program End ************************/