Timus Online Judge1009 - - K-based Numbers
4667 단어 dp
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 valid K based numbers, containing N 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. Sample input output
2 10
90
dp[i][j]i 비트, i위 j의 합법적인 수
/************************************************************************* > File Name: TOJ1009.cpp > Author: ALex > Mail: [email protected] > Created Time: 2015 05 14 18 55 28 ************************************************************************/
#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>
using namespace std;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
LL dp[20][20];
int main() {
int k, n;
while (~scanf("%d%d", &n, &k)) {
memset(dp, 0, sizeof(dp));
for (int i = 0; i < k; ++i) {
dp[1][i] = 1;
}
for (int i = 2; i <= n; ++i) {
for (int j = 1; j < k; ++j) {
dp[i][0] += dp[i - 1][j];
}
for (int j = 1; j < k; ++j) {
for (int l = 0; l < k; ++l) {
dp[i][j] += dp[i - 1][l];
}
}
}
LL ans = 0;
for (int i = 1; i < k; ++i) {
ans += dp[n][i];
}
printf("%lld
", ans);
}
return 0;
}