1103. Integer Factorization (30)

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".

Sample Input 1:

169 5 2

Sample Output 1:

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

Sample Input 2:

169 167 3

Sample Output 2:

Impossible

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按题目要求分解成k个数的p次方的和，要求和最大。用dfs加剪枝来实现。第五个case比较耗时间，要尽量减少时间，比如不能用自带的pow函数，这个比较耗时间。

代码：

#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <vector>#include <cmath>#include <algorithm>using namespace std;int n,k,p;vector<int>res;vector<int>mypow;int max_idx;void init_pow(){for(int i=0;;i++){int num=1;for(int j=0;j<p;j++){num*=i;}mypow.push_back(num);max_idx=i;if(num>=400) return;}}int Max=0;void dfs(int target,int count,int high,vector<int>fac){if(count==0){if(target==0){if(res.empty()) res=fac;else{int sum=0;for(int i=0;i<k;i++){sum+=fac[i];}if(sum>Max) {Max=sum;res=fac;}}}return;}if(target<count) return;while(mypow[high]>target-count+1) high--;for(int i=high;mypow[i]>=target/count;i--){if(target>=mypow[i]){fac[count-1]=i;dfs(target-mypow[i],count-1,i,fac);}}}int main(){cin>>n>>k>>p;init_pow();vector<int>fac(k);dfs(n,k,max_idx,fac);if(!res.empty()){reverse(res.begin(),res.end());printf("%d = %d^%d",n,res[0],p);for(int i=1;i<res.size();i++){printf(" + %d^%d",res[i],p);}}else {printf("Impossible");}}