hdu 1545 01-K Code

01-K Code

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 406    Accepted Submission(s): 96

Problem Description

Consider a code string S of N symbols, each symbol can only be 0 or 1. In any consecutive substring of S, the number of 0's differs from the number of 1's by at most K. How many such valid code strings S are there? 

For example, when N is 4, and K is 3, there are two invalid codes: 0000 and 1111. 

 

Input

The input consists of several test cases. For each case, there are two positive integers N and K in a line.

N is in the range of [1, 62].

K is in the range of [2, 5].

 

Output

Output the number of valid code strings in a line for each case.

 

Sample Input

1 2
4 3

 

Sample Output

2
14

 

分析：dp[i][j][k] 表示遍历到第i位(从1开始)，当前0-1的最小值为j，0-1的最大值为k，可知j的最大值为0，k的最小值为0；

            dp[0][0][0]表示空串

转移方程：dp[i+1][min(j+1,0)][k+1]+=dp[i][j][k];//第i+1位添0

                    dp[i+1][j-1][max(k-1,0)]+=dp[i][j][k];//第i+1位添1

由于j可能为负数，故加一个偏移值5

#include<cstdio>
#include<cstring>
#define ll __int64
const int W=5;
int n,k;
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
ll dp[63][10][10];
int main(){
	int i,j,p;
	while(~scanf("%d%d",&n,&k)){
		memset(dp,0,sizeof(dp));
		dp[0][W][W]=1;
		for(i=0;i<n;i++){
			for(j=-k+W;j<=W;j++)for(p=W;p<=k+W;p++){
				dp[i+1][min(j+1,W)][p+1]+=dp[i][j][p];
				dp[i+1][j-1][max(p-1,W)]+=dp[i][j][p];
			}
		}
		ll ans=0;
		for(i=-k+W;i<=W;i++)for(j=W;j<=k+W;j++)ans+=dp[n][i][j];
		printf("%I64d\n",ans);
	}
	return 0;
}