1103. Integer Factorization (30)

1.根据题目的特点，n的最大值不超过400，可以使用深度遍历搜索答案

2.加入限制条件进行剪枝，如i的p次方不能大于n-k+1，因为剩下的数至少均为1，剩下还需要k-1个数需要进行分配，每个数至少分配为1，共k-1，所以i的p次方最大只能为n-（k-1）=n-k+1，而最大值不能超过上一次分配的。

//#include<string>//#include <iomanip>#include<vector>#include <algorithm>//#include<stack>#include<set>#include<queue>#include<map>//#include<unordered_set>#include<unordered_map>//#include <sstream>//#include "func.h"//#include <list>#include<stdio.h>#include<iostream>#include<string>#include<memory.h>#include<limits.h>using namespace std;void dfs(int n, int count, int p,int preNum, vector<int>&ans,int ansSum, vector<pair<int,vector<int>>>&totalAns, vector<vector<int>>&num){if (n == 0 && count == 0){totalAns.push_back({ansSum, ans});}else if ((n == 0 && count != 0) || (n != 0 && count == 0))return;else{for (int i = min(n - count+1,preNum); i >= 1; i--){//进行剪枝if (num[i][p - 1] != -1 && num[i][p - 1]<=n-count+1){ans.push_back(i);ansSum += i;dfs(n - num[i][p - 1], count - 1, p,i, ans,ansSum, totalAns, num);ansSum -= i;ans.pop_back();}}}}bool cmp(const pair<int, vector<int>>&a, const pair<int, vector<int>>&b){if (a.first > b.first)return true;else if (a.first == b.first){for (int i = 0; i < a.second.size(); i++){if (a.second[i] > b.second[i]) return true;else if (a.second[i] < b.second[i]) return false;}}else return false;}int main(void){int n, k, p;cin >> n >> k >> p;vector<vector<int>> num(401, vector<int>(7, 0));for (int i = 1; i < 401; i++){num[i][0] = i;for (int j = 1; j < 7; j++){if (num[i][j - 1] == -1) num[i][j] = -1;else{num[i][j] = num[i][j - 1] * i;if (num[i][j] > 400) num[i][j] = -1;}}}vector<int> ans(0);vector<pair<int,vector<int>>> totalAns(0);int ansSum = 0;dfs(n, k, p,n, ans,ansSum, totalAns, num);sort(totalAns.begin(), totalAns.end(), cmp);if (totalAns.size() != 0){cout << n << " = ";for (int i = 0; i < totalAns[0].second.size(); i++){cout << totalAns[0].second[i] << "^" << p;if (i != totalAns[0].second.size() - 1){cout << " + ";}}cout << endl;}else {cout << "Impossible" << endl;}return 0;}