PAT A1103—DFS

 Integer Factorization

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：给定正整数N,K,P,将N表示成K个正整数（可以相同，递减排列）的P次方的和，即N=n1^p+...+nk^p。如果有多种方案，那么选择底数和n1+...+nk最大的方案：如果还有多种方案，那么选择底数序列的字典序最大的方案。

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
int n,k,p,maxFacSum = -1;   //n,k,p如题所诉，maxFacSum记录最大底数之和
vector<int> fac,ans,temp;   //fac记录0^p,1^p...i^p,ans存放最优底数序列，temp存放递归中的临时底数序列

int power(int x) {  //power函数计算x^p
    int ans = 1;
    for(int i=0; i<p; i++){
        ans *= x;
    }
    return ans;
}

void init() {
    int i=0, temp=0;
    while(temp <= n) {  //当i^p没有超过n时，不断把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; //更新最大底数之和
        }
        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);   //"选"的分支
        temp.pop_back();    //"选"的分支结束后把刚加进去的数pop掉
        DFS(index-1, nowK, sum, facSum);    //"不选"的分支
    }
}

int main() {
    scanf("%d%d%d",&n,&k,&p);
    init(); //初始化fac数组
    DFS(fac.size()-1,0,0,0);    //从fac的最后一位开始往前搜索
    if(maxFacSum == -1) printf("Impossible
");
    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;
}