PAT(甲级)1103

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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

AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;
const int SIZE = 404;
int a[SIZE];

int pow(int base,int exp){
	int res =1;
    while(exp){
    	if(exp & 0x1)
    	    res *= base;
    	base *= base;
    	exp >>= 0x1; 	    
    }
    return res;	
}

int init_a(const int &n,const int &p){
	a[0] = 0;
	for(int i=1;i<=n;i++){
		int tmp = pow(i,p);
		if(tmp <= n)
		    a[i] = tmp;
		else
		    return i-1;
	}
	return n;  
}

vector<int> res,vtmp;
int numbers=0;

void display(){
	vector<int>::iterator iter = vtmp.begin();
	while(iter != vtmp.end())
	    cout <<*iter++ <<' ';
	cout <<endl;
}
/*
void find_factors(int n,int k,int p,int start,int end,int sum,int cnt){
	bool flag = true;
	numbers++;
	for(;start<=end && flag;start++){
		if(sum == n && cnt == k)
		    res = vtmp;
		else if(sum < n && cnt < k){
            int tmp = a[start];
    	    vtmp.push_back(start);	
		    find_factors(n,k,p,start,end,sum+tmp,cnt+1);
			vtmp.pop_back();	
		}else if(sum > n)
		    flag = false;
    }		
}
*/
void find_factors(int n,int k,int p,int start,int end,int sum,int cnt){
	bool flag = true;
//	numbers++;
	for(;start<=end && flag;start++){
		int tmpsum = sum +a[start];
		vtmp.push_back(start);
		if(tmpsum == n && cnt+1 == k)
		    res = vtmp;
		else if(tmpsum < n && cnt+1 < k){
		    find_factors(n,k,p,start,end,tmpsum,cnt+1);	
		}else if(tmpsum > n)
		    flag = false;
		vtmp.pop_back();
    }		
}
//*/


int main()
{
	int N,K,P;
	scanf("%d%d%d",&N,&K,&P);
	int end = init_a(N,P);
    find_factors(N,K,P,1,end,0,0);
    if(!res.empty()){
    	printf("%d = %d^%d",N,res[K-1],P);
    	for(int i=K-2;i>=0;i--){
    		printf(" + %d^%d",res[i],P);
    	}
    	printf("\n");
    }else
        printf("Impossible\n");
//    cout <<numbers <<endl;
    return 0;
}

下面的版本代码运行的更快，但有一个bug始终找不出来，望高手指出：

<pre name="code" class="cpp">#include <cstdio>
#include <iostream>
#include <algorithm>
#include <vector>


//where is the bug???
///
using namespace std;
const int SIZE = 404;
int a[SIZE];

int pow(int base,int exp){
	int res =1;
    while(exp){
    	if(exp & 0x1)
    	    res *= base;
    	base *= base;
    	exp >>= 0x1; 	    
    }
    return res;	
}

int init_a(const int &n,const int &p){
	a[0] = 0;
	for(int i=1;i<=n;i++){
		int tmp = pow(i,p);
		if(tmp <= n)
		    a[i] = tmp;
		else
		    return i-1;
	}
	return n;  
}

vector<int> res,vtmp;
int sum=0,vsum=0;

void display(){
	vector<int>::iterator iter = vtmp.begin();
	while(iter != vtmp.end())
	    cout <<*iter++ <<' ';
	cout <<endl;
}
int numbers = 0;
//bug version/

void find_factors(int n,int k,int p,int start){
	numbers++;
	if(k ==0 && n == 0){
		res = vtmp;
//		display();
	}else if(k > 0 && n > 0){
	    for(int i= start;i>=1;i--){
	       	int tmp = a[i];
	    	vtmp.push_back(i);
 	    	find_factors(n-tmp,k-1,p,i);
	    	vtmp.pop_back();
       }		
	}
}

/*
void find_factors(int n,int k,int p,int start){
//	numbers++;
	for(int i= start;i>=1;i--){
		int newn = n - a[i];
		vtmp.push_back(i);
		if(newn == 0 && k == 1)
		    res = vtmp;
		else if(newn >0 && k > 1)
		    find_factors(newn,k-1,p,i);
		vtmp.pop_back();
    }		

}
*/

int main()
{
	int N,K,P;
	scanf("%d%d%d",&N,&K,&P);
	int start = init_a(N,P);
    find_factors(N,K,P,start);
    if(!res.empty()){
    	printf("%d = %d^%d",N,res[0],P);
    	for(int i=1;i<K;i++){
    		printf(" + %d^%d",res[i],P);
    	}
    	printf("\n");
    }else
        printf("Impossible\n");
//    cout << numbers <<endl;
    return 0;
}