SRM 622 D2L3: Subsets, math, backtrack

 

 

The number of non-1 elements in a set that meet the condition is very small. You can use backtracing and pruning to run quickly.

 

Code:

 

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                      using namespace std;#define CHECKTIME() printf(%.2lf, (double)clock() / CLOCKS_PER_SEC)typedef pair
                      
                        pii;typedef long long llong;typedef pair
                       
                         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 
                        
                          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 
                         
                           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 ************************/