PAT 1103. Integer Factorization (30)(dfs)
2018-03-07 21:10
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题意:给你n,k,p,让你求一个长度为k的数组a满足, n = a1^p + a2^p + .... ak^p, 如果有多个解,则取a的和最大的,如果还是有多解,则取字典序最大的。(n <= 400, k <= n, 1 < p <=7)
思路:因为数据量很小,所以可以爆搜,考虑有多解的情况需要选择和最大其次字典序最大,所以我们可以倒着搜。
代码:#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 25;
const int maxm = 505;
int a[maxm], res[maxm], fac[maxn];
int n, p, k, tmpSum;
bool ok;
void dfs(int cur, int num, int sum, int val)
{
if(sum > n) return ;
if(num == k)
{
if(sum == n && val > tmpSum)
{
ok = 1;
tmpSum = val;
for(int i = 0; i < k; i++)
res[i] = a[i];
}
else return ;
}
for(int i = cur; i >= 1; i--)
{
if(sum+fac[i] <= n)
{
a[num] = i;
dfs(i, num+1, sum+fac[i], val+i);
}
}
}
int main(void)
{
while(cin >> n >> k >> p)
{
for(int i = 1; i <= 20; i++)
{
int tmp = 1;
for(int j = 1; j <= p; j++)
tmp = tmp*i;
fac[i] = tmp;
}
ok = tmpSum = 0;
dfs(20, 0, 0, 0);
if(ok)
{
printf("%d = ", n);
for(int i = 0; i < k; i++)
{
printf("%d^%d", res[i], p);
if(i != k-1) printf(" + ");
}
puts("");
}
else puts("Impossible");
}
return 0;
}
思路:因为数据量很小,所以可以爆搜,考虑有多解的情况需要选择和最大其次字典序最大,所以我们可以倒着搜。
代码:#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 25;
const int maxm = 505;
int a[maxm], res[maxm], fac[maxn];
int n, p, k, tmpSum;
bool ok;
void dfs(int cur, int num, int sum, int val)
{
if(sum > n) return ;
if(num == k)
{
if(sum == n && val > tmpSum)
{
ok = 1;
tmpSum = val;
for(int i = 0; i < k; i++)
res[i] = a[i];
}
else return ;
}
for(int i = cur; i >= 1; i--)
{
if(sum+fac[i] <= n)
{
a[num] = i;
dfs(i, num+1, sum+fac[i], val+i);
}
}
}
int main(void)
{
while(cin >> n >> k >> p)
{
for(int i = 1; i <= 20; i++)
{
int tmp = 1;
for(int j = 1; j <= p; j++)
tmp = tmp*i;
fac[i] = tmp;
}
ok = tmpSum = 0;
dfs(20, 0, 0, 0);
if(ok)
{
printf("%d = ", n);
for(int i = 0; i < k; i++)
{
printf("%d^%d", res[i], p);
if(i != k-1) printf(" + ");
}
puts("");
}
else puts("Impossible");
}
return 0;
}
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