Lightoj 1009. K-based Numbers(DP)

1009. K-based Numbers

Time limit: 1.0 second
Memory limit: 64 MB

Let’s consider K-based numbers, containing exactlyN digits. We define a number to be valid if itsK-based notation doesn’t contain two successive zeros. For example:

• 1010230 is a valid 7-digit number;
• 1000198 is not a valid number;
• 0001235 is not a 7-digit number, it is a 4-digit number.

Given two numbers N and K, you are to calculate an amount of validK based numbers, containingN digits.

You may assume that 2 ≤ K ≤ 10; N ≥ 2;N +K ≤ 18.

Input

The numbers N and K in decimal notation separated by the line break.

Output

The result in decimal notation.

搜索一次就可以了，记录一下当前位置的数字还有长度

/*********************************************** * Author: fisty * Created Time: 2015/6/10 10:26:53 * File Name   : 1009.cpp *********************************************** */#include <iostream>#include <cstring>#include <deque>#include <cmath>#include <queue>#include <stack>#include <list>#include <map>#include <set>#include <string>#include <vector>#include <cstdio>#include <bitset>#include <algorithm>using namespace std;#define Debug(x) cout << #x << " " << x <<endl#define Memset(x, a) memset(x, a, sizeof(x))const int INF = 0x3f3f3f3f;typedef long long LL;typedef pair<int, int> P;#define FOR(i, a, b) for(int i = a;i < b; i++)int n, k;  int ans;void dfs(int cur, int m){    //cur: 现在这一位 m:长度    if(m == n) { ans++; return ;}    for(int i = 0;i < k; i++){        if(cur == 0 && i == 0) continue;  //当前一位与下一位不能同时为0        dfs(i, m+1);    }}int main() {    //freopen("in.cpp", "r", stdin);    cin.tie(0);    ios::sync_with_stdio(false);    cin >> n >> k;        ans = 0;    for(int i = 1;i < k; i++){        dfs(i, 1);    }    cout << ans << endl;    return 0;}

还有就是写成递推的形式，第i位只有0和非0

定义dp[i][0]为第i位为0，dp[i][1] 为第i位非0

则  dp[i][0] = dp[i-1][1];

      dp[i][1] = (k-1)*(dp[i][1-1] + dp[i-1][0])     (2 <= i <= n)

/*********************************************** * Author: fisty * Created Time: 2015/6/10 10:26:53 * File Name   : 1009.cpp *********************************************** */#include <iostream>#include <cstring>#include <deque>#include <cmath>#include <queue>#include <stack>#include <list>#include <map>#include <set>#include <string>#include <vector>#include <cstdio>#include <bitset>#include <algorithm>using namespace std;#define Debug(x) cout << #x << " " << x <<endl#define Memset(x, a) memset(x, a, sizeof(x))const int INF = 0x3f3f3f3f;typedef long long LL;typedef pair<int, int> P;#define FOR(i, a, b) for(int i = a;i < b; i++)int n, k;  int ans;LL dp[20][2];int main() {    //freopen("in.cpp", "r", stdin);    cin.tie(0);    ios::sync_with_stdio(false);    cin >> n >> k;        dp[1][0] = 0; dp[1][1] = k-1;    for(int i = 2;i <= n; i++){        dp[i][1] = (k-1) * (dp[i-1][1] + dp[i-1][0]);        dp[i][0] = dp[i-1][1];    }    cout << dp[n][1] + dp[n][0] << endl;    return 0;}