PAT(甲级)1103

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;}