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递推DP URAL 1009 K-based Numbers

2015-05-06 17:38 375 查看
题目传送门

题意:n位数,k进制,求个数
分析:dp[i][j] 表示i位数,当前数字为j的个数;若j==0,不加dp[i-1][0];

代码1:

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

const int MAXN = 22;
const int INF = 0x3f3f3f3f;
long long dp[MAXN][MAXN];

int main(void)        //URAL 1009 K-based Numbers
{
//freopen ("B.in", "r", stdin);

int n, k;
while (scanf ("%d%d", &n, &k) == 2)
{
memset (dp, 0, sizeof (dp));
for (int i=1; i<k; ++i)    dp[1][i] = 1;
for (int i=2; i<=n; ++i)
{
for (int j=0; j<k; ++j)
{
if (!j)
for (int l=1; l<k; ++l)    dp[i][j] += dp[i-1][l];
else
for (int l=0; l<k; ++l)    dp[i][j] += dp[i-1][l];
}
}

long long ans = 0;
for (int i=0; i<k; ++i)    ans += dp
[i];
printf ("%I64d\n", ans);
}

return 0;
}


代码2(空间优化):

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

const int MAXN = 22;
const int INF = 0x3f3f3f3f;
long long dp[MAXN];

int main(void)        //URAL 1009 K-based Numbers
{
//freopen ("B.in", "r", stdin);

int n, k;
while (scanf ("%d%d", &n, &k) == 2)
{
memset (dp, 0, sizeof (dp));

dp[0] = 1;    dp[1] = k - 1;
for (int i=2; i<=n; ++i)
{
dp[i] = (dp[i-1] + dp[i-2]) * (k-1);
}

printf ("%I64d\n", dp
);
}

return 0;
}
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