Tight words
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 2079 | Accepted: 1041 |
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.
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
4 1 2 5 3 5 8 7
Sample Output
100.00000 40.74074 17.38281 0.10130 题意就是给你一个从0到k的字母表,让你输出长度为n的由前面给你的字母表组成的符合Tight Words的字符串占左右字符串的百分比
动态规划的子结构划分还是有点不太清楚,看到题目,不知道从哪里下手,DP还得多练,多体会,DP无处不在,DP才是王道
Memory: 400K | Time: 0MS | |
Language: GCC | Result: Accepted |
#include<stdio.h> double opt[10][101];//opt[i][j]表示已i结尾的长度为j的tight字符串个数 int main() { int k,n; int i,j; double sum; freopen("input","r",stdin); while(scanf("%d %d",&k,&n)!=EOF) { memset(opt,0,sizeof(opt)); sum=0; for(i=1;i<=k+1;i++) opt[i][1]=1;//所有长度为1的字符串本身就是tight字符串 for(j=2;j<=n;j++)//opt[i][j]=opt[i-1][j-1]+opt[i][j-1]+opt[i+1][j-1] 状态转移方程 { for(i=1;i<=k+1;i++) { opt[i][j]=opt[i-1][j-1]+opt[i][j-1]+opt[i+1][j-1]; } } for(i=0;i<=k+1;i++)//计算字符长度为n的字符串个数 { sum+=opt[i][n]; } printf("%.5f/n",sum/pow(k+1,n)*100); } return 0; }