PAT(Advanced Level)1103. Integer Factorization

1103. Integer Factorization (30)

时间限制

1200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 = n₁^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

    一个简单的深搜，当时做天梯训练的时候做了一下，没做出来，感觉好坑的题，回来重新敲一遍就过了，谁知道当初我做了啥。

#include <bits/stdc++.h>using namespace std;int n,m,k,flag = 0,ans = 0;int a[500],b[500];void dfs(int d,int sum,int cnt,int t){    if(sum>n)return;    if(cnt==m){        if(sum==n){        flag = 1;        if(t>ans){            ans = t;            for(int i = 0;i < m;i++)                b[i] = a[i];        }        }        return;    }    for(int i = d;i >= 1;i--){            a[cnt] = i;            dfs(i,sum+pow(i,k),cnt+1,t+i);    }}int main(){    cin>>n>>m>>k;    int temp = sqrt(n);    dfs(temp,0,0,0);    if(!flag)cout<<"Impossible"<<endl;    else {        cout<<n<<" = ";        for(int i = 0;i < m;i++){            cout<<b[i]<<"^"<<k;            if(i!=m-1){                cout<<" + ";            }        }    }    return 0;}