【PAT】1103 Integer Factorization(30 分)
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 (≤), K (≤) and P (1). The numbers in a line are separated by a space.
Output Specification:
For each case, if the solution exists, output in the format:
N = n[1]^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 1, or 1, 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 { , } is said to be larger than { , } if there exists 1 such that ai=bi for i<L and aL>bL.
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
C++代码如下:
#include<iostream>
#include<cmath>
#include<vector>
using namespace std; int n, k, p;
vector<int>v;
vector<int>temp,ans;
int sum_g=;
void init() {
int t;
for (int i = ; i < ; i++) {
t = pow(i, p);
if (t <= n)
v.push_back(t);
else break;
}
}
void DFS(int index, int sum, int nowk, int sumk) {
if ( nowk == k && sum == n) {
if (sumk > sum_g) {
sum_g = sumk;
ans = temp;
}
return;
}
if ( sum>n || nowk > k)return;
if (index - >= ) {
temp.push_back(index);
DFS(index , sum + v[index], nowk + , sumk + index);
temp.pop_back();
DFS(index - , sum, nowk, sumk);
}
}
int main() {
cin >> n >> k >> p;
init();
DFS(v.size() - , , , );
if (ans.size() > ) {
cout << n << " = ";
for (int i = ; i < ans.size() - ; i++)
cout << ans[i] << "^" << p << " + ";
cout << ans[ans.size() - ] << "^" << p << endl;
}
else
cout << "Impossible" << endl;
return ;
}
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