Time Limit: 1sec Memory Limit:256MB
Description
Given is an alphabet {0, 1, ... , k}, 0 <= k <= 9 . We say that a word of length n over this alphabet is tight if any two neighbour digits in the word do not differ by more than 1. Your task is to find out the percentage
of tight words of length n over the given alphabet.
Input
Input is a sequence of lines, each line contains two integer numbers k and n, 1 <= n <= 100.
Output
For each line of input, output the percentage of tight words of length n over the alphabet {0, 1, ... , k} with 5 fractional digits.
Sample Input
Sample Output
100.00000
40.74074
17.38281
0.10130
題意是用0到k這k+1個符號，組成一個n位的數，如果這個n位數相鄰的位置上的數字差都不超過1，那麼這個n位數就稱爲tight words，最後問tight words的個數佔總數的百分比，因爲總數有(k+1)^n之多，所以只能用double來存儲了，我們用dp[i][j]表示i位數，最高位上是j的tight words的個數，初始化後，很容易就可以遞推出來，因爲個數比較大，long long也存不下，只能使用double了。
這道題目簡單的數位dp，狀態轉移很簡單，處理好邊界就可以了。
// Problem#: 11599
// Submission#: 3667449
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <cmath>
#include <ctime>
#include <cctype>
#include <climits>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <sstream>
#include <iostream>
#include <algorithm>
double dp[111][111];
int main(int argc, char *argv[]){
int k, n;
int N=111;
while (scanf("%d%d",&k,&n)!=EOF){
for (int i = 0; i < N; ++i){
for (int j = 0; j < N; ++j){
dp[i][j] = 1.0;
}
}
for (int i = 2; i<N; ++i){
for (int j = 0; j<= k;++j){
dp[i][j] = dp[i-1][j];
if (j > 0){
dp[i][j] += dp[i-1][j-1];
}
if (j < k) {
dp[i][j] += dp[i-1][j+1];
}
}
}
double ans = 0;
double base = pow(k + 1, n);
for (int i = 0; i <= k; ++i) {
ans += dp[n][i];
}
printf("%.5f\n", ans * 100.0 / base);
}
return 0;
}
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