1103 Integer Factorization (30)（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 (<=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 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 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


题意比较好懂，容易想到dfs，但是不能纯回溯，需要剪枝，或者说按照一个非递减的顺序去试每个值，而且判断结果时满足k项就更新，因为从第一层开始，往下dfs，每一层的初始试验值都大于等于上一层的正在试验值，如果遇到跟之前一次结果相同的，序列一定比之前大。
代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
int n,k,p;
vector<int> temp,ans;
int pow_[401];///把可能用到的p次方都算出来，要多算一个 这样循环结束 不至于死循环（如果算的pow_[j] 都小于n 那么最大的max(j) + 1对应pow_[j]初始为0
int pow(int t) {
    int d = 1;
    for(int i = 0;i < p;i ++) {
        d *= t;
    }
    return d;
}
void dfs(int t,int s,int last) {///last是上一层正在尝试的  ，本次从last开始 达到非递减的目的
    if(t >= k) {
        if(!s) {
            ans = temp;
        }
        return;
    }
    while(pow_[last] <= s) {///这里防止死循环  如果last = max(j) 保证pow_[last]能使循环结束
        temp.push_back(last);
        dfs(t + 1,s - pow_[last],last);
        temp.pop_back();
        last ++;
    }
}
int main() {
    scanf("%d%d%d",&n,&k,&p);
    int j = 1,d = 1;
    while(d <= n) {
        pow_[j ++] = d;
        d = pow(j);
    }
    pow_[j] = d;///多算一次 不然判断会死循环
    dfs(0,n,1);
    if(ans.empty())printf("Impossible");
    else {
        printf("%d = %d^%d",n,ans[k - 1],p);
        for(int i = k - 2;i >= 0;i --) {
            printf(" + %d^%d",ans[i],p);
        }
    }
}