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01-K Code
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 728 Accepted Submission(s): 209
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
Sample Output
2
14
//有一个长为 i 的 0-1 组成的序列，问任意连续长的子序列 0 1 数量差都不超过 m 的情况多少种
// dp[i][j][k] 意思是长为 i 的序列的子序列中 0 比 1 最多多 j 个，且长为 i 的序列的子序列中 1 比 0 最多多 k 个
所以 i，j 最小为 0
dp [i][j][k] = dp[i-1][max(0,j-1)][k+1] // 填 1
+dp[i-1][j+1][max(0,k-1)] // 填 0
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define LL long long 4 5 int n,m; 6 LL dp[105][10][10]; //长为 i 的序列，0比1最多多 j 个，且1比0最多多 k 个 7 8 int main() 9 {10 while (scanf("%d%d",&n,&m)!=EOF)11 {12 memset(dp,0,sizeof(dp));13 dp[0][0][0]=1;14 for (int i=0;i<=n;i++)15 {16 for (int j=0;j<=m;j++)17 {18 for (int k=0;k<=m;k++)19 {20 dp[i+1][j+1][max(k-1,0)]+=dp[i][j][k]; // 下一个数是 021 dp[i+1][max(j-1,0)][k+1]+=dp[i][j][k]; // 下一个数是 122 }23 }24 }25 LL ans = 0;26 for (int i=0;i<=m;i++)27 for (int j=0;j<=m;j++)28 ans += dp[n][i][j];29 printf("%lld\n",ans);30 }31 return 0;32 }
01-K Code