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 = 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 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,⋯,aK } is said to be larger than { b1,b2,⋯,bK } if there exists 1≤L≤K 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
一， 我的思想
DFS()函数只有一个参数，就是当前的P次方的和，其他的参数：
1，当前下标：用开头的for搜索，为了使后一个比前一个小，通过全局变量num存储vector末尾元素，for从num开始往后循环。
2，当前矢量数组中元素个数：不设这个变量，直接在if中判定vec.size()与K的大小关系。
3，当前数组中元素和，在搜索到符合条件的数组时候在与全局变量比较。
缺点：不好维护，不模块化。
二，正确思想：
把这三个参数和当前的P次方的和都添加到DFS()的形参中，同时把P次方到N的那些元素和结果用vector存起来，元素是下标，结果是数组中的元素。
三，小tip:
1，把一个vector的值赋给另一个vector不能用
for (int i = 0; i < K; i++) {answer[i] = temp[i];}而应当用
answer = temp;直接相等即可。
2，不要被例子带着走，像判断nowK与K的关系，应当是nowK > K，而当初是nowK > 5，还有其他的几个参数用了例子中的常量，切记，不要被例子带着走。
四，我的代码
#include<cstdio>
#include<vector>
#include<math.h>
using namespace std;
//到17:05
int N = 0;
int K = 0;
int P = 0;
int sum_1 = -1;
int num = 0;
int countting = 0;
int arr[500] = { 0 };
int simu[500] = { 0 };
vector<int> vec;
void compare(int arr_1[]) {
int sum = 0;
int flag = 0;
for (int i = 0; i < K; i++) {
sum += arr_1[i];
}
if (sum > sum_1) {
sum_1 = sum;
for (int i = 0; i < K; i++) {
arr[i] = arr_1[i];
}
}
num = arr[0];
}
void DFS(int sum) {
for (int i = num; i > 0; i--) {
int P_mul = 1;
for (int j = 0; j < P; j++) {
P_mul *= i;
}
if ((vec.size() < K) && (sum + P_mul <= N)) {
vec.push_back(i);
num = i;
DFS(sum + P_mul);
if (vec.size() == K && sum + P_mul == N) {
countting++;
vector<int>::iterator it = vec.begin();
for (int k = 0; k < K; k++) {
simu[k] = *(it + k);
}
compare(simu);
}
vec.pop_back();
}
}
}
int main() {
scanf("%d %d %d", &N, &K, &P);
num = pow(N, 1.0 / P);
DFS(0);
if (countting == 0) {
printf("Impossible");
return 0;
}
printf("%d = ", N);
for (int i = 0; i < K; i++) {
printf("%d^2", arr[i]);
if (i != K - 1) {
printf(" + ");
}
}
return 0;
}五，正确代码：
#include<cstdio>
#include<vector>
#include<math.h>
using namespace std;
//到17:05
int N = 0, K = 0, P = 0;
int maxSum = -1;
vector<int> fac, answer, temp;
int pow(int num) {
int ans = 1;
for (int i = 0; i < P; i++) {
ans *= num;
}
return ans;
}
void init() {
int temp = 0;
int i = 0;
while (temp<=N) {
fac.push_back(temp);
temp = pow(++i);
}
}
void DFS(int index, int nowK, int sum, int facsum) {
if (sum == N&&nowK == K) {
if (facsum > maxSum) {
maxSum = facsum;
answer = temp;//for (int i = 0; i < K; i++) {answer[i] = temp[i];}这样写是不对的
}
return;
}
if (nowK > K || sum > N)return; //当初>=不对 当初nowK > 5不对
if(index - 1 >= 0) {
temp.push_back(index);
DFS(index, nowK + 1, sum + fac[index], facsum + index);
temp.pop_back();
DFS(index - 1, nowK, sum, facsum);
}
}
int main() {
scanf("%d %d %d", &N, &K, &P);
// int num = pow(N, 1.0 / P);
init();
DFS(fac.size()-1,0,0,0);
if (maxSum == -1) {
printf("Impossible");
return 0;
}
printf("%d = ", N);
for (int i = 0; i < K; i++) {
printf("%d^%d", answer[i],P);
if (i != K - 1) {
printf(" + ");
}
}
return 0;
}