1103. Integer Factorization (30) -- dfs 回溯 待解决
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 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
http://www.patest.cn/contests/pat-a-practise/1103
// 21分 各种运行超时
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
#define N 20
int n , k , p;
bool flag ;
vector<int> ans ;
int numP(int num) // num的p次方
{
int sums = num ;
for(int i = 1 ; i < p ; i++)
{
sums *= num ;
}
return sums;
}
int sumv(vector<int> vv) // 求和
{
int sums = 0 ;
for(int i = k-1 ;i >= 0 ; i --)
{
sums += vv[i];
}
return sums;
}
bool cmp(vector<int> v1,vector<int> v2)
{
if(sumv(v1) > sumv(v2))
return true;
else if (sumv(v1) == sumv(v2))
{
for(int i = k-1 ;i >= 0 ; i --)
{
if(v1[i] > v2[i])
return true;
if(v1[i] < v2[i])
return false;
}
}
return false;
}
vector<int> ansT ;
bool dfs(int n , int step)
{
int i;
if(n < 0 || step > k)
return false;
if(step == k && n == 0)
{
if(flag == false)
{
flag = true;
for(i = 0 ;i < k;i++)
{
ans.push_back(ansT[i]);
}
}else{
if( cmp(ansT,ans) )
{
ans.clear();
for(i = 0 ;i < k;i++)
{
ans.push_back(ansT[i]);
}
}
}
return true;
}else{
for(i = 1 ;i <= n ; i++)
{
int num = numP(i);
ansT.push_back(i);
dfs(n-num , step+1);
if((int)ansT.size() > 0 )
ansT.pop_back();
}
}
return false;
}
int main()
{
//freopen("in.txt","r",stdin);
int i;
while(scanf("%d%d%d",&n,&k,&p) != EOF)
{
int tmpn = n ;
ans.clear();
ansT.clear();
flag = false;
dfs(n,0);
if(flag)
{
printf("%d = %d^%d",tmpn,ans[k-1],p);
for(i = k-2;i>=0;i--)
{
printf(" + %d^%d",ans[i],p);
}
}else{
printf("Impossible");
}
printf("\n");
}
return 0 ;
}
// 27分代码
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
#define N 20
int n , k , p;
int ansSum;
vector<int> ans ;
int numP(int num) // num的p次方
{
int sums = num ;
for(int i = 1 ; i < p ; i++)
{
sums *= num ;
}
return sums;
}
int sumv(vector<int> vv) // 求最后结果的和
{
int sums = 0 ;
for(int i = k-1 ;i >= 0 ; i --)
{
sums += vv[i];
}
return sums;
}
bool cmp(vector<int> v1,vector<int> v2) // 比较结果的大小
{
for(int i = k-1 ;i >= 0 ; i --)
{
if(v1[i] > v2[i])
return true;
if(v1[i] < v2[i])
return false;
}
return false;
}
vector<int> ansT ;
bool dfs(int n , int step)
{
int i;
if(n < 0 || step > k)
return false;
if(step == k && n != 0)
return false;
if(step == k && n == 0)
{
if(ansSum == -1)
{
ansSum = 0 ;
for(i = 0 ;i < k;i++)
{
ans.push_back(ansT[i]);
ansSum += ansT[i];
}
}else{
int sumT = sumv(ansT);
if( sumT > ansSum ) // 和 大
{
ansSum = sumT ;
ans.clear();
for(i = 0 ;i < k;i++)
{
ans.push_back(ansT[i]);
}
}else if(sumT == ansSum && cmp(ansT,ans)) // 和 相等
{
ans.clear();
for(i = 0 ;i < k;i++)
{
ans.push_back(ansT[i]);
}
}
}
return true;
}else{
for(i = 1 ;i <= n ; i++)
{
int num = numP(i);
if(num > n) // 这里比n大 则直接 return
return false;
int len = (int)ansT.size();
if(len > 0 && num < ansT[len-1]) // 跳过该元素 保证后一个元素比前一个元素大于等于
continue;
ansT.push_back(i);
dfs(n-num , step+1); // dfs 下一个
if((int)ansT.size() > 0) // 回溯 pop出去
ansT.pop_back();
}
}
return false;
}
int main()
{
//freopen("in.txt","r",stdin);
int i;
while(scanf("%d%d%d",&n,&k,&p) != EOF)
{
int tmpn = n ;
ans.clear();
ansT.clear();
ansSum = -1 ;
dfs(n,0);
if(ansSum != -1)
{
printf("%d = %d^%d",tmpn,ans[k-1],p);
for(i = k-2;i>=0;i--)
{
printf(" + %d^%d",ans[i],p);
}
}else{
printf("Impossible");
}
printf("\n");
}
return 0 ;
}// ac代码
#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
int num[21];
int maxSum = 0;
vector<int> res;
vector<int> vec;
/** start 表示开始的数 k 为个数 n 为要求的数 sum 为当前已经求得的和 */
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){
// 显然 n < num[i] 就不满足了
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(); // 回溯 直接弹出最后一个数
}
}
}
}
int main(void)
{
int n, k, p;
scanf("%d%d%d", &n, &k, &p);
// num[i] 表示 i^p 这样就直接不用pow了
for (int i = 1; i < 21; ++i){
num[i] = 1;
for (int j = 0; j < p; ++j){
num[i] *= i;
}
}
// dfs(int start, int k, int n, int sum)
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;
} ac代码学习：http://blog.csdn.net/apie_czx/article/details/48415197