pat1103

1. 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小于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 + … nK^P

where ni (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, … aK } is said to be larger than { b1, b2, … bK } if there exists 1<=L<=K such that ai=bi for ibL

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

一看就是dfs，主要考的是你的dfs的方式，还有剪枝；
此处声明：用math库里的pow会超时第五个样例点
我原先错的dfs方式每次因子都从1开始遍历，这样会造成太多的重复情况，像169里有一种情况是6，6，6，6，5但也会遍历5，6 ，6，6，6多次，像下面代码似的总是从小到大，少了太多太多的遍历

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<stack>#include<cmath>#include<vector>using namespace std;int n,k,p;vector<int>que;vector<int>ppx;int MAX = 0;int Pow(int m,int n){    int sum = 1;    for(int i = 1;i <= n;++i)    {        sum *= m;    }    return sum;}void dfs(int m){    //printf("%d\n",m);    if(m < 0){        return ;    }    if(que.size() > k){        return ;    }    if(m != 0 && que.size() == k)        return ;    if(m == 0 && que.size() == k)    {        int sum = 0;        for(int i = 0;i < k;++i)        {            sum += que[i];        }        if(sum >= MAX){            ppx = que;            MAX = sum;        }        return ;    }    //这儿的优化总是找的是从小到大的    //后面更新了的虽然和相同，但也比原先的大，    //这种方式减少了很多重复情况    int start = (que.size() == 0) ? 1:que[que.size() - 1];    //每次因子都从前一个因子往大走    for(int i = start;i <= (int)sqrt(m);++i)    {        if(Pow(i,p) <= m)        {            que.push_back(i);            dfs(m - pow(i,p));            que.pop_back();        }    }}int main(){    scanf("%d %d %d",&n,&k,&p);    dfs(n);    if(ppx.size() == 0){        printf("Impossible\n");        return 0;    }    printf("%d = %d^%d",n,ppx[ppx.size() - 1],p);    for(int i = ppx.size() - 2;i >= 0;--i)    {        printf(" + %d^%d",ppx[i],p);    }    cout << endl;    return 0;}