PAT - 甲级 - 1103. Integer Factorization (30)（dfs回溯+减枝）

题目描述：

The K-P factorization of a positive integer N is to write N as the sum of the P-th power of K positive integers. You are supposed to write a program to find the K-P factorization of N for any positive integers N, K and P.

Input Specification:

Each input file contains one test case which gives in a line the three positive integers N (<=400), K (<=N) and P (1<P<=7). The numbers in a line are separated by a space.

Output Specification:

For each case, if the solution exists, output in the format:

N = n₁^P + ... n_K^P

where n_i (i=1, ... K) is the i-th factor. All the factors must be printed in non-increasing order.

Note: the solution may not be unique. For example, the 5-2 factorization of 169 has 9 solutions, such as 12² + 4² + 2² + 2² + 1², or 11² + 6² + 2² + 2² + 2², or more. You must output the one with the maximum sum of the factors. If there is a tie, the largest factor sequence must be chosen -- sequence { a₁, a₂, ... a_K } is said to belarger than { b₁, b₂, ... b_K } if there exists 1<=L<=K such that a_i=b_i for i<L and a_L>b_L

If there is no solution, simple output "Impossible".

169 5 2

169 = 6^2 + 6^2 + 6^2 + 6^2 + 5^2

169 167 3

Impossible

题目思路：

递归搜索出所有的情况（注意减枝），输出数字和最大的一组即可。

题目代码：

#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;
int n, k, p;
int ans[405];
int sum = 0,cnt = 0;
vector<int>v[100000];  
void dfs(int cur, int num){// 退出条件 if(sum > n || cur > k) return ;	if(sum == n){  if(cur == k){ // 符合条件的结果			for(int i = 0; i < cur; i++){v[cnt].push_back(ans[i]);}		cnt++;return ;}}// 递归范围 for(int i = num; i >=1 ;i--){int temp = 1;for(int j = 0; j < p; j++){temp *= i;}ans[cur] = i;sum += temp;dfs(cur+1, i);sum -= temp; // 回溯	}	
}
int main(){scanf("%d%d%d",&n,&k,&p);dfs(0,sqrt(n));// 如果符合条件的结果数为0, impossible if(cnt == 0){printf("Impossible\n"); }else{// 记录和最大的一组答案的编号 int maxid, maxsum = -1;for(int i = 0; i < cnt; i++){int cursum = 0;for(int j = 0; j <v[i].size(); j++){cursum += v[i][j];}if(cursum > maxsum){maxsum = cursum;maxid = i;} }// 答案输出 printf("%d = ",n);for(int i = 0; i < v[maxid].size(); i++){if(i)printf(" + ");printf("%d^%d",v[maxid][i],p);}}	return 0;
}