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
#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<string>
#include<vector>
#include<queue>
#include<numeric>//include accumulate
using namespace std;
int aceil[8] = { 0, 0, 20, 7, 4, 3, 2, 2 };//表示上限，比如当p=4时，ceil[4]=4，表示4^4<400而5^4>400
//这个4就是ans中每个数字的上限，ans[i]必须都<=aceil[p]，因为ceil已经是关键字了，所以我这里用了aceil
int n, k, p;//放在全局变量中，这样DFS中就可以少一些参数传递
vector<int>v_tmp;//存放可能的答案的序列
vector<vector<int>>ans;//存放所有的答案
int kpow(int x, int y){
int ret = x;
while (--y){
ret *= x;
}
return ret;
}
void DFS(int start, int curSum, int count){//分别表示开始的数字，现在的sum和计数
for (int i = start; i <= aceil[p]; ++i){
int sum_tmp = curSum + kpow(i, p);//*point，这里用pow测试点5会超时，需要用自己写的kpow
if (sum_tmp > n)return;//剪枝，和大了要剪掉
if (count == k - 1 && sum_tmp < n)continue;//剪枝，到达个数了和小了也剪掉
if(count == k - 1 && sum_tmp == n){
v_tmp.push_back(i);
ans.push_back(v_tmp);
v_tmp.pop_back();
return;
}
v_tmp.push_back(i);//典型的DFS模式：挖坑，跳入，再填坑（自创）
DFS(i, sum_tmp, count + 1);
v_tmp.pop_back();
}
return;
}
/*如果不用accumulate或者不熟悉，就可以自己写一个，效果一样的
int vSum(vector<int>Vx){
int ret = 0;
for (int i = 0; i < Vx.size(); ++i){
ret += Vx[i];
}
return ret;
}
*/
bool cmp(vector<int>V1, vector<int>V2){
if (accumulate(V1.begin(), V1.end(), 0) != accumulate(V2.begin(), V2.end(), 0))//accumulate第3个参数是init，初值设置为0就OK
return accumulate(V1.begin(), V1.end(), 0) > accumulate(V2.begin(), V2.end(), 0);
else{
return V1 > V2;//vector自带的<运算符，可以按元素依次比较，符合我们的要求
}
}
void print(vector<int> Vx){//对于vector的格式化输出，再定制更改一下
cout << n << " = ";
for (int i = 0; i < Vx.size(); ++i){
if (i) cout << " + " << Vx[i] << "^" << p;
else cout << Vx[i]<< "^" << p;
}
}
int main(){
freopen("F://Temp/input.txt", "r", stdin);
cin >> n >> k >> p;
/*题目中说p > 1，不过如果p可以等于1的话，应该是如下处理
if (p == 1){//当p为1时，就可以直接做处理
v_tmp.push_back(n - k + 1);
for (int i = 0; i < k - 1; i++){
v_tmp.push_back(1);
}
print(v_tmp);
cout << endl;
return 0;
}
*/
DFS(1, 0, 0);//初始值
if (ans.size() == 0){//如果没有答案，直接输出返回
cout << "Impossible" << endl;
return 0;
}
for (int i = 0; i < ans.size(); ++i){
reverse(ans[i].begin(), ans[i].end());//因为答案是按递增的，为了比较和输出方便，倒置一下
}
sort(ans.begin(), ans.end(), cmp);//对多个答案按照题意进行排序，把最佳的放在最前面
print(ans[0]);
cout << endl;
return 0;
}
#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
int num[21];
int maxSum = 0;
vector<int> res;
vector<int> vec;
void dfs(int start, int k, int n, int sum){
if (k == 0){
if (n == 0){
if (sum >= maxSum){
res = vec;//只用记录最后一组答案就好，因为从数学的角度看，这是最佳的
maxSum = sum;
}
}
}
else{
if (n > 0){
for (int i = start; i < 21; ++i){
if(n - num[i] < 0)//剪枝，没有这个判断测试点5过不了
break;
vec.push_back(i);
dfs(i, k - 1, n - num[i], sum + i);
vec.pop_back();
}
}
}
}
int main(void){
int n, k, p;
scanf("%d%d%d", &n, &k, &p);
for (int i = 1; i < 21; ++i){//这样就直接不用pow了
num[i] = 1;
for (int j = 0; j < p; ++j){
num[i] *= i;
}
}
dfs(1, k, n, 0);
if (res.empty())
printf("Impossible\n");
else{
printf("%d = ", n);
printf("%d^%d", res[k - 1], p);
for (int i = k - 2; i >= 0; --i){
printf(" + ");
printf("%d^%d", res[i], p);
}
printf("\n");
}
return 0;
}
代码更简洁，也更好了。并且：