PAT甲级——1103 Integer Factorization (DFS)

本文同步发布在CSDN：https://blog.csdn.net/weixin_44385565/article/details/90574720

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 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

题目大意：将一个正整数N分解成K个正整数的P次方和，在多个结果里面找出因子之和最大的，若因子之和相同，字典序大的为答案。

思路：主要是DFS的思想，建立一个数组F，用来储存 1~m的P次方，m^P为≤N的最大正整数。find()里面传入四个变量，n为当前find()里面的for循环次数；cnt初始值为K，cnt=0作为递归的边界；tmpSum储存因子之和；sum是总和，sum=N才是符合条件的备选答案~

下一层的递归里的n总是小于等于上一层递归里的n，所以保证了字典序，不需要画蛇添足地写compera函数来筛选答案了（一开始就是因为这个操作导致测试点2答案错误），若无必要，勿增操作。

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
int N, K, P, m, fSum = -1;
vector <int> ans, F, tmpA;
void find(int n,int cnt, int tmpSum, int sum);

int main()
{
    scanf("%d%d%d", &N, &K, &P);
    int i = 1;
    F.push_back(0);
    while (1) {
        int x = pow(i, P);
        if (x > N)
            break;
        else {
            F.push_back(x);
            i++;
        }
    }
    m = F.size() - 1;
    find(m, K, 0, 0);
    if (ans.empty()) {
        printf("Impossible
");
        return 0;
    }
    printf("%d =", N);
    for (int i = 0; i < K; i++) {
        printf(" %d^%d", ans[i], P);
        if (i < K - 1) {
            printf(" +");
        }
    }
    printf("
");
    return 0;
}
void find(int n, int cnt, int tmpSum, int sum) {
    if(n==0) return;
    if (cnt == 0) {
        if (fSum < tmpSum) {
            if (sum == N) {
                ans = tmpA;
                fSum = tmpSum;
            }
        }
        return;
    }
    for (int i = n; i > 0; i--) {
        if (sum <= N) {
            tmpA.push_back(i);
            find(i, cnt - 1, tmpSum + i, sum + F[i]);
            tmpA.pop_back();
        }
    }
}