PAT-A1103 Integer Factorization 题目内容及题解

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

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 12​^2​​+4^​2​​+2^​2​​+2^​2​​+1^​2​​, or 11​^2​​+6^​2​​+2​^2​​+2^​2​​+2^​2​​, 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 { a​1​​,a​2​​,⋯,a​K​​ } is said to be larger than { b​1​​,b​2​​,⋯,b​K​​ } if there exists 1≤L≤K such that a​i​​=b​i​​ for ib​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

题目给出正整数N、个数K和次数P，要求将正整数N分解为K个正整数P次方的和，并按照题目要求输出。

解题思路

1. 读入数据，并提前打好表存储P次方数字；
2. 通过深度优先遍历做背包问题，找到符合要求的答案，期间注意剪枝以节约时间；
3. 按照题目要求的格式输出答案并返回零值。

代码

#include
#include
#include
using namespace std;
vector vec,temp,ans;
int N,K,P,maxsum=-1;

void Init(){
    int i=1,num=0;
    scanf("%d%d%d",&N,&K,&P);
    while(num<=N){
        vec.push_back(num);
        num=pow(i,P);
        i++; 
    }
    temp.resize(K);
}

void DFS(int current,int sum,int depth,int facsum){
    int i;
    if(sum>N){
        return;
    }//超过范围，剪枝
    if(depth==K){
        if(sum==N&&facsum>maxsum){
            ans=temp;
            maxsum=facsum;
        }
        return;
    }
    for(i=current;i>0;i--){
        if(sum+vec[i]<=N){
            temp[depth]=i;
            DFS(i,sum+vec[i],depth+1,facsum+i);
        }
    }
}

void Print(){
    int i;
    if(maxsum<0){
        printf("Impossible\n");
    }else{
        printf("%d = ",N);
        for(i=0;i

运行结果