HDU 1058 Humble Numbers 动态规划 或 暴力
2014-08-25 01:17
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题目大意:
就是求第n个满足可以由任意个2,3,5,7相乘能得到的正整数
注意输出时的格式即可
大致思路:
这题首先明显能暴力枚举=_=,这是一个方法
好一点的可以知道用dp[ i ] 表示第 i 个满足条件的数是 dp[ i ] 那么dp[ i ] 肯定来自于前面的数乘上2或者3,5,7这样注意一下更新条件即可
代码如下:
Result : Accepted Memory : 308 KB Time : 250 ms
/*
* Author: Gatevin
* Created Time: 2014/8/9 20:03:37
* File Name: hehe.cpp
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;
lint dp[5845];
int n;
int main()
{
memset(dp, 0, sizeof(dp));
for(int i = 1; i <= 10; i++) dp[i] = i;
for(int i = 11; i <= 5842; i++)
{
dp[i] = 2*dp[i - 1];
for(int j = i - 2; j >= 1; j--)
{
if(dp[j] * 7 <= dp[i - 1]) break;
else dp[i] = min(dp[i], dp[j] * 7);
if(dp[j] * 5 <= dp[i - 1]) continue;
else dp[i] = min(dp[i], dp[j] * 5);
if(dp[j] * 3 <= dp[i - 1]) continue;
else dp[i] = min(dp[i], dp[j] * 3);
if(dp[j] * 2 <= dp[i - 1]) continue;
else dp[i] = min(dp[i], dp[j] * 2);
}
}
char s1, s2;
while(scanf("%d", &n) && n)
{
switch(n % 10)
{
case 1 : s1 = 's'; s2 = 't'; break;
case 2 : s1 = 'n'; s2 = 'd'; break;
case 3 : s1 = 'r'; s2 = 'd'; break;
default : s1 = 't';s2 = 'h';
}
if(n % 100 == 11 || n % 100 == 12 || n % 100 == 13)
{
s1 = 't';
s2 = 'h';
}
printf("The %d%c%c humble number is %I64d.\n", n, s1, s2, dp
);
}
return 0;
}
就是求第n个满足可以由任意个2,3,5,7相乘能得到的正整数
注意输出时的格式即可
大致思路:
这题首先明显能暴力枚举=_=,这是一个方法
好一点的可以知道用dp[ i ] 表示第 i 个满足条件的数是 dp[ i ] 那么dp[ i ] 肯定来自于前面的数乘上2或者3,5,7这样注意一下更新条件即可
代码如下:
Result : Accepted Memory : 308 KB Time : 250 ms
/*
* Author: Gatevin
* Created Time: 2014/8/9 20:03:37
* File Name: hehe.cpp
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;
lint dp[5845];
int n;
int main()
{
memset(dp, 0, sizeof(dp));
for(int i = 1; i <= 10; i++) dp[i] = i;
for(int i = 11; i <= 5842; i++)
{
dp[i] = 2*dp[i - 1];
for(int j = i - 2; j >= 1; j--)
{
if(dp[j] * 7 <= dp[i - 1]) break;
else dp[i] = min(dp[i], dp[j] * 7);
if(dp[j] * 5 <= dp[i - 1]) continue;
else dp[i] = min(dp[i], dp[j] * 5);
if(dp[j] * 3 <= dp[i - 1]) continue;
else dp[i] = min(dp[i], dp[j] * 3);
if(dp[j] * 2 <= dp[i - 1]) continue;
else dp[i] = min(dp[i], dp[j] * 2);
}
}
char s1, s2;
while(scanf("%d", &n) && n)
{
switch(n % 10)
{
case 1 : s1 = 's'; s2 = 't'; break;
case 2 : s1 = 'n'; s2 = 'd'; break;
case 3 : s1 = 'r'; s2 = 'd'; break;
default : s1 = 't';s2 = 'h';
}
if(n % 100 == 11 || n % 100 == 12 || n % 100 == 13)
{
s1 = 't';
s2 = 'h';
}
printf("The %d%c%c humble number is %I64d.\n", n, s1, s2, dp
);
}
return 0;
}
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