文章目录
题目原文Input Specification:Output Specification:Sample Input 1:Sample Output 1:Sample Input 2:Sample Output 2:
生词如下:题目大意:代码如下:核心代码(模板)柳神的代码:
强烈推荐,刷PTA的朋友都认识一下柳神–PTA解法大佬
本文由参考于柳神博客写成
柳神的博客,这个可以搜索文章
柳神的个人博客,这个没有广告,但是不能搜索
还有就是非常非常有用的 算法笔记 全名是
算法笔记 上级训练实战指南 //这本都是PTA的题解
算法笔记
PS 今天也要加油鸭
题目原文
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,⋯,a**K } is said to be larger 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
生词如下:
PS:有事朦胧题,但是这题如果我不翻译的话,肯定是做不出来的.我只知道这题可以套模板,但是并不知道,这题的变量都是啥.
Factorization 因子分解
题目大意:
给你一个数N 然后一个数K,一个数P
要你吧N因数分解成K个数的p次方的和
而且要求这K个数的和是最大的
这是一道非常经典的DFS算法.
我们采用的方法就是套用模板
代码如下:
#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;
//n ,k,p如题所示,maxFacsum就是最大底数之和
int n, k, p, maxFacSum = -1;
//fac记录0^p 1^p ..... i^p是的i^p为不超过n的最大数
//ans存放最优底数序列,temp存放递归中的临时底数序列
vector<int> fac, ans, temp;
//power函数计算x^p
int power(int x) {
int ans = 1;
for (int i = 0; i < p; ++i) {
ans *= x;
}
return ans;
}
//init函数预处理fac数组,注意吧0也存进去.
void init() {
int i = 0,temp = 0;
while (temp <= n) { //当i^p没有超过n时,不断吧i^p加入fac
fac.push_back(temp);
temp = power(++i);
}
}
//DFS函数,当前访问fac[index],nowK为当前选中个数
//sum为当前选中的数之和,facSum为当前选中的底数之和
void DFS(int index, int nowK, int sum, int facSum) {
if (sum == n && nowK == k) { //找到一个满足的序列
if (facSum > maxFacSum) { //底数之和更优
ans = temp; //更新最优底数序列
maxFacSum = facSum;
}
return;
}
if (sum > n || nowK > k) return; //这种情况下不会产生答案,直接返回
if (index - 1 >= 0) { //fac[0]不需要选择
temp.push_back(index); //把底数index加入临时序列temp
DFS(index, nowK + 1, sum + fac[index], facSum + index);
temp.pop_back();
DFS(index-1, nowK, sum, facSum);//不选的分支
}
}
int main(void) {
scanf("%d%d%d", &n, &k, &p);
init(); //初始化init数组
DFS(fac.size() - 1, 0, 0, 0); //从fac的最后以为开始往前搜索
if (maxFacSum == -1) printf("Impossible\n");
else {
printf("%d = %d^%d", n, ans[0],p); //输出ans的结果
for (int i = 1; i < ans.size(); ++i)
printf(" + %d^%d", ans[i], p);
}
return 0;
}
核心代码(模板)
void DFS(int index, int nowK, int sum, int facSum) {
if (sum == n && nowK == k) { //找到一个满足的序列
if (facSum > maxFacSum) { //底数之和更优
ans = temp; //更新最优底数序列
maxFacSum = facSum;
}
return;
}
if (sum > n || nowK > k) return; //这种情况下不会产生答案,直接返回
if (index - 1 >= 0) { //fac[0]不需要选择
temp.push_back(index); //把底数index加入临时序列temp
DFS(index, nowK + 1, sum + fac[index], facSum + index);
temp.pop_back();
DFS(index-1, nowK, sum, facSum);//不选的分支
}
}
柳神的代码:
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
int n, k, p, maxFacSum = -1;
vector<int> v, ans, tempAns;
void init() {
int temp = 0, index = 1;
while (temp <= n) {
v.push_back(temp);
temp = pow(index, p);
index++;
}
}
void dfs(int index, int tempSum, int tempK, int facSum) {
if (tempK == k) {
if (tempSum == n && facSum > maxFacSum) {
ans = tempAns;
maxFacSum = facSum;
}
return;
}
while(index >= 1) {
if (tempSum + v[index] <= n) {
tempAns[tempK] = index;
dfs(index, tempSum + v[index], tempK + 1, facSum + index);
}
if (index == 1) return;
index--;
}
}
int main() {
scanf("%d%d%d", &n, &k, &p);
init();
tempAns.resize(k);
dfs(v.size() - 1, 0, 0, 0);
if (maxFacSum == -1) {
printf("Impossible");
return 0;
}
printf("%d = ", n);
for (int i = 0; i < ans.size(); i++) {
if (i != 0) printf(" + ");
printf("%d^%d", ans[i], p);
}
return 0;
}
