Time limit: 3.000 seconds
Given is an alphabet {0, 1, ... , k}, 0 <= k <= 9 . We say that a word of length n over this alphabet is tightif any two neighbour digits in the word do not differ by more than 1.
Input is a sequence of lines, each line contains two integer numbers k and n, 1 <= n <= 100. 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
4 1 2 5 3 5 8 7
Output for the sample input
100.00000 40.74074 17.38281 0.10130
题意：给定两个数k,n。
用 {0, 1, ... , k}的数组成一个n个数的序列。假设这个序
列每两个相邻的数相差<=1,就记为是tight，求这样的序列占总序列的比率。
思路： dp[i][j]表示第i为数字是j的概率 。
即 dp[i][j] = 1/(k+1) * (dp[i-1][j-1]+dp[i-1][j] + dp[i+1][j] );
注意下边界就OK了。
#include
#include
#include
using namespace std;
const int maxn=105;
double dp[maxn][15],t,ans;
int n,k;
void initial()
{
memset(dp,0,sizeof(dp));
t=1.0/(k+1),ans=0.0;
for(int j=0; j<=k; j++) dp[1][j]=t;
}
void solve()
{
for(int i=2; i<=n; i++)
for(int j=0; j<=k; j++)
{
dp[i][j]+=t*dp[i-1][j];
if(j!=0) dp[i][j]+=t*dp[i-1][j-1];
if(j!=k) dp[i][j]+=t*dp[i-1][j+1];
}
for(int j=0; j<=k; j++) ans+=dp[n][j];
printf("%.5lf\n",ans*100);
}
int main()
{
while(scanf("%d %d",&k,&n)!=EOF)
{
initial();
solve();
}
return 0;
}