1103. Integer Factorization (30)

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

#include<cstdio>
#include<cstring>
#include<algorithm>
#define bug(x) printf("%d****\n",x)
typedef long long ll;
using namespace std;
/*
用 long long 要不然会炸，再就是注意剪枝 
*/

const int maxn=410;
ll max_num;
ll ed[maxn],cnt;
ll fina[maxn];

ll N,K,P;
ll pw(ll base,int n){
	ll ans=1;
	for(int i=1;i<=n;i++)
		ans=ans*base;
	return ans;
}

ll flag,sum2;
void check(ll* ed,ll *fina){
	ll sum1=0;
	for(int i=1;i<=K;i++)
		sum1+=ed[i];
	if(sum1>sum2){
		sum2=sum1;
		for(int i=1;i<=K;i++)
			fina[i]=ed[i];
	}
	else if(sum1==sum2){
		int tmp=0;
		for(int i=1;i<=K;i++){
			if(ed[i]>fina[i]){
				tmp=1;
				break;
			}
			else if(ed[i]<fina[i]){
				break;
			}
		}
		if(tmp){
			for(int i=1;i<=K;i++)
				fina[i]=ed[i];
		}
	}
}

void dfs(int now,ll sum,int pre){
	//printf("****now:%d sum:%d\n",now,sum);
	if(now==K+1&&sum==N){
		if(!flag){
			flag=1;
			sum2=0;
			for(int i=1;i<=K;i++){
				fina[i]=ed[i];
				sum2+=ed[i];
			}	
		}
		else{
			check(ed,fina);
		}
		return;
	}
	if(sum+(K-now+1)*pw(pre,P)<N)//一个剪枝，我们的时间就可以直接缩短，切记 
		return; 
	if(sum>N||now>K)return;
	for(ll i=pre;i>=1;i--){
		ed[now]=i;
		dfs(now+1,sum+pw(i,P),i);
	}
}

int main(){
	flag=0;
	scanf("%lld %lld %lld",&N,&K,&P);
	max_num=N-(K-1);
	dfs(1,0,max_num);
	if(flag){
		printf("%lld = ",N);
		for(int i=1;i<=K;i++){
			if(i==1)
				printf("%lld^%lld",fina[i],P);
			else	
				printf(" + %lld^%lld",fina[i],P);
		}
		printf("\n");
	}
	else
		printf("Impossible\n");
	return 0;
}