【PAT甲级】1103 Integer Factorization（30 分）

题目链接

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

思路：DFS+回溯

代码：

#include <iostream>
#include <sstream>
#include <string>
#include <cstdio>
#include <queue>
#include <cstring>
#include <map>
#include <cstdlib>
#include <stack>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define mem(a,n) memset(a,n,sizeof(a))
#define DBGS() cout<<"START\n"
#define DBGE() cout<<"END\n"
const int N = 80+5;
const ll INF = 0x3f3f3f3f;
const double eps=1e-4;

int n,k,p;
int maxFacSum=-1;
vector<int>fac,ans,tmpAns;
int myPow(int n) {
    int ans=1;
    for(int i=0; i<p; i++) {
        ans*=n;
    }
    return ans;
}
void init() {
    int ttmp=0,index=1;
    while(ttmp<=n) {
        fac.push_back(ttmp);
//        printf("%d ",ttmp);
        ttmp=myPow(index);
        index++;
    }
}
/***
index:因数分解数组fac的下标
tmpSum:分解序列和
tmpK:因子分解序列的个数
facSum:因子底数之和
*/
void dfs(int index,int tmpSum,int tmpK,int facSum) {
    if(tmpSum>n||tmpK>k)
        return;
    if(tmpK==k) {
        if(tmpSum==n&&facSum>maxFacSum) {
            ans=tmpAns;
            maxFacSum=facSum;
        }
        return ;//剪枝
    }
    if(index>=1) {
        tmpAns.push_back(index);///选择
        dfs(index,tmpSum+fac[index],tmpK+1,facSum+index);
        tmpAns.pop_back();///不选
        dfs(index-1,tmpSum,tmpK,facSum);
    }
}
int main() {
    scanf("%d%d%d",&n,&k,&p);
    init();
    tmpAns.reserve(k);
    dfs(fac.size()-1,0,0,0);
//    printf("%d %d %d\n",fac.size(),ans.size(),maxFacSum);
    if(maxFacSum==-1) {
        printf("Impossible");
        return 0;
    }
    printf("%d = ",n);
    for(int i=0; i<ans.size(); i++) {
        if(i)
            printf(" + ");
        printf("%d^%d",ans[i],p);
    }
    return 0;
}