PAT 1103. Integer Factorization (30)
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 + ... 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 i<L and aL>bL
If there is no solution, simple output "Impossible".
Sample Input 1:169 5 2Sample Output 1:
169 = 6^2 + 6^2 + 6^2 + 6^2 + 5^2Sample Input 2:
169 167 3Sample Output 2:
Impossible
题意看一下数据应该能看懂
因为数据比较小，所以直接dfs暴力搜索 + 剪枝 就行了
dfs(int start, int x, int step) //start为此时的系数， x为N剩下的值， step为系数的个数
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cmath>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
const double PI = acos(-1.0);
using namespace std;
#define esp 1e-8
int n, k, p;
int a[500];
int b[500], tot, temp;
int power(int a, int b)
{
int ans = 1;
while(b)
{
if(b & 1)
ans = a * ans;
b >>= 1;
a = a * a;
}
return ans;
}
void dfs(int start, int x, int step)
{
x -= power(start, p);
b[step ++] = start;
if(step == k && x == 0)
{
int s = 0;
for(int i = 0; i < step; ++ i)
{
s += b[i];
}
if(s >= a[k]) //如果系数和比前面的大就替换掉
{
for(int i = 0; i < step; ++ i)
{
a[i] = b[i];
}
temp = 1;
}
a[k] = s;把和保存在a[k]
return;
}
if(x <= 0 || step >= k)
{
return ;
}
for(int i = start; power(i, p) <= x; ++ i)
{
if((i * (k - step)) > x) //注意剪枝
break;
dfs(i, x, step);
}
return;
}
int main()
{
int i, j;
while(~scanf("%d%d%d", &n, &k, &p))
{
temp = 0;
for(i = 1; power(i, p) <= n; ++ i)
{
dfs(i, n, 0);
}
if(temp == 0)
printf("Impossible\n");
else
{
int mm = 0;
int res = 0;
sort(a, a + k + 1);
printf("%d =", n);
for(i = k - 1; i >= 0; -- i)
{
if(i != k - 1)
printf(" +");
printf(" %d^%d", a[i], p);
}
printf("\n");
}
}
}