1103 Integer Factorization (30分)
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 = n[1]^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,?,*aK* } is said to be larger** than { b1,b2,?,*b**K* } 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
分析
将正整数N分解为K项数字(因子)的P次方之和,因子以非增顺序排列。多个解中取因子之和最大的,若和相同则取因子的字典序在前的。没解输出Impossible
。
- 设置数组a:将index=0开始的数字的P次方存到数组a中,直到a[index]>n。
- 全局变量:设置数组tempAns和ans,变量facSum和maxFacSum
数组tempAns存放当前解的临时因子们,ans存放最优解的因子们。
maxFacSum最优解的因子之和,初值设置为-1。
- 开始dfs:
- tempK表示当前K的个数,当tempK==K时,判断tempSum当前总和==N且(是否为最优解)facSum当前解的因子之和>maxFacSum:更新最优结果变量ans和maxFacSum。
- index从高后前遍历数组a
- 判断当前项a[index]是否加入解:tempSum+a[index]<=N,满足则更新tempAns,并递归调用dfs确定下一项
- 判断是否完成一个解:index==1
- 若不满足tempSum+a[index]<=N,则index--,因子变小再判断。
代码(参考柳神)
#include<iostream>
#include<vector>
#include<cmath>
using namespace std;
int K,N,P;
vector<int> a,tempAns,ans;
int maxFacSum=-1;
/*初始化:计算a数组*/
void init() {
int temp=0;
for(int i=1; temp<=N; i++) {//注意;i从1开始
a.push_back(temp);
temp=pow(i,P);
}
}
/*dfs*/
void dfs(int index,int tempK,int tempSum,int facSum) {
if(tempK==K) {
if(tempSum==N&&facSum>maxFacSum) {
ans=tempAns;
maxFacSum=facSum;
}
return;
}
while(index>=1) {
if(tempSum+a[index]<=N) {
tempAns[tempK]=index;//使用数组访问必须对vector进行resize
dfs(index,tempK+1,tempSum+a[index],facSum+index);
}
if(index==1) return;
index--;
}
}
int main() {
scanf("%d %d %d",&N,&K,&P);
init();
tempAns.resize(K);
dfs(a.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;
}
单词
- factorization:因式分解
factor:因子
- If there is a
tie
, ..:平手(相等)
原文地址:https://www.cnblogs.com/musecho/p/12236349.html