PAT 1103. Integer Factorization (30) DFS+树状数组。还是人家的代码好

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 + ... 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 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

所谓DFS，也不过是k重for循环而已。算是最笨的办法，一组一组数据试的，应该有更好的方法。直接k重for循环，有一组会超时。改成树状数组就过了。

#include <iostream>
#include<stdio.h>
#include<string>
#include<queue>
#include<algorithm>
#include<string.h>
#include<vector>
#include<map>
#include<math.h>
using namespace std;
int n,k,p;
int tree_array[500]={0};
int last[500]={0};
int summ(int x)
{
int a=0;
while(x>0)
{
a+=tree_array[x];
x-=x&(-x);
}
return a;
}

void updata(int x,int y)
{
while(x<401)
{
tree_array[x]+=y;
x+=x&(-x);
}
}

struct ZZ
{
int s[500];
int sum;
bool operator < (const ZZ &A) const
{
if(sum!=A.sum) return sum>A.sum;
for(int i=1;i<=p;i++)
if(s[i]!=A.s[i])
return s[i]>A.s[i];
}

}node[5];

int edge;
vector<ZZ> Q;

int cc(int x)
{
int i=1;
for(int s=0;s<p;s++)
i*=x;
return i;
}
int lav=2;
void dfs(int ceng)
{
int lala;
if(ceng==1) lala=edge;
else lala=node[0].s[ceng-1];
for(int i=1;i<=lala;i++)
{
node[0].s[ceng]=i;
if(ceng==5&&i>4)
lav++;
updata(ceng,cc(i)-last[ceng]);
last[ceng]=cc(i);
int sum=0;
//for(int j=1;j<=ceng;j++)      //用这个会超时，改用树状数组。
// sum+=cc(node[0].s[j]);
sum=summ(ceng);
if(sum>n) break;
// if(sum==n&&ceng==k)               //测试输出
// {
//    for(int i=1;i<=ceng;i++)
//      cout<<node[0].s[i]<<' ';
//     cout<<endl;
// }
if(ceng<k)
dfs(ceng+1);
else
{
if(sum==n)
{
node[0].sum=0;
for(int bb=1;bb<=k;bb++)
node[0].sum+=node[0].s[bb];
Q.push_back(node[0]);
}
}
}
}

int main()
{

cin>>n>>k>>p;
int x=1;
while(x<=n)    //这里必须是小于等于，不能是小于
{
edge++;
x=1;
for(int i=0;i<p;i++)
x*=edge;
}
edge--;

dfs(1);

if(Q.empty()) cout<<"Impossible";
else
{
sort(Q.begin(),Q.end());
cout<<n<<" = ";
cout<<Q[0].s[1]<<'^'<<p;
for(int i=2;i<=k;i++)
cout<<" + "<<Q[0].s[i]<<'^'<<p;
// printf(" +  %d^%d",Q[0].s[i],p);
}
return 0;
}

看了下别人的代码，原理一模一样，不过人家的优雅轻便了不知道多少。。羞愧

分析：dfs深度优先搜索。先把i从0开始所有的i的p次方的值存储在v[i]中，直到v[i]
> n为止。然后深度优先搜索，记录当前正在相加的index（即v[i]的i的值），当前的总和tempSum，当前K的总个数tempK，以及因为题目中要求输出因子的和最大的那个，所以保存一个facSum为当前因子的和，让它和maxFacSum比较，如果比maxFacSum大就更新maxFacSum和要求的ans数组的值。
在ans数组里面存储因子的序列，tempAns为当前深度优先遍历而来的序列，从v[i]的最后一个index开始一直到index
== 1，因为这样才能保证ans和tempAns数组里面保存的是从大到小的因子的顺序。一开始maxFacSum == -1，如果dfs后maxFacSum并没有被更新，还是-1，那么就输出Impossible，否则输出答案。

#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;
int n, k, p, maxFacSum = -1;
vector<int> v, ans, tempAns;
void init() {
int temp = 0, index = 1;
while(temp <= n) {
v.push_back(temp);
temp = pow(index, p);
index++;
}
}
void dfs(int index, int tempSum, int tempK, int facSum) {
if(tempSum == n && tempK == k) {
if(facSum > maxFacSum) {
ans = tempAns;
maxFacSum = facSum;
}
return ;
}
if(tempSum > n || tempK > k) return ;
if(index >= 1) {
tempAns.push_back(index);
dfs(index, tempSum + v[index], tempK + 1, facSum + index);
tempAns.pop_back();
dfs(index - 1, tempSum, tempK, facSum);
}
}
int main() {
scanf("%d%d%d", &n, &k, &p);
init();
dfs(v.size() - 1, 0, 0, 0);
if(maxFacSum == -1) {
printf("Impossible");
return 0;
}
printf("%d = ", n);
for(int i = 0; i < ans.size(); i++) {
if(i != 0) printf(" + ");
printf("%d^%d", ans[i], p);
}
return 0;
}