SRM 622 D2L3: Subsets, math, backtrack

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

符合条件的集中非1的元素个数是很少的，可以用回溯加剪枝，实际运行速度很快。

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){// addint 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 addif (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 ************************/

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