Adjacent Bit Counts(uvalive)
For a string of n bits x1, x2, x3,…, xn, the adjacent bit count of the string (AdjBC(x)) is given by
x1 ∗ x2 + x2 ∗ x3 + x3 ∗ x4 + . . . + xn−1 ∗ xn
which counts the number of times a 1 bit is adjacent to another 1 bit. For example:
AdjBC(011101101) = 3
AdjBC(111101101) = 4
AdjBC(010101010) = 0
Write a program which takes as input integers n and k and returns the number of bit strings x of n bits (out of 2 n ) that satisfy AdjBC(x) = k. For example, for 5 bit strings, there are 6 ways of getting AdjBC(x) = 2:
11100, 01110, 00111, 10111, 11101, 11011
Input
The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. Each data set is a single line that contains the data set number, followed by a space, followed by a decimal integer giving the number (n) of bits in the bit strings, followed by a single space, followed by a decimal integer (k) giving the desired adjacent bit count. The number of bits (n) will not be greater than 100 and the parameters n and k will be chosen so that the result will fit in a signed 32-bit integer.
Output
For each data set there is one line of output. It contains the data set number followed by a single space, followed by the number of n-bit strings with adjacent bit count equal to k.
Sample Input
10
1 5 2
2 20 8
3 30 17
4 40 24
5 50 37
6 60 52
7 70 59
8 80 73
9 90 84
10 100 90
Sample Output
1 6
2 63426
3 1861225
4 168212501
5 44874764
6 160916
7 22937308
8 99167
9 15476
10 23076518
题解:求一个长度为p,构造一个价值为n的字符串的方法数;
首先如果我们要构造一个价值为7的字符串,若分为两部分的话,我们有1+6,2+5,3+4等3种方案,对于1+6,我们又可以分为一个子问题,将6分为两部分,以此类推;
这样就满足一个最优子结构;这样我们把原问题可以分为若干子问题,我们用的dp[i][j]表示长度为i-1,价值为j的字符串的方法数,由于最后一位0与1两种情况,我们可以定义3维数组;
dp[i][j][0]=dp[i-1][j][0]+dp[i-1][j][1];
dp[i][j][1]=dp[i-1][j][0]+dp[i-1][j-1][1];
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 typedef long long ll; 6 const int MAXN=1e3+10; 7 int m,n; 8 int dp[MAXN][MAXN][2]; 9 using namespace std; 10 int main() 11 { 12 cin>>m; 13 int p,k,u; 14 while(m--) 15 { 16 cin>>p>>k>>u; 17 memset(dp,0,sizeof(dp)); 18 dp[1][0][1]=1; 19 dp[1][0][0]=1; 20 for(int i=2;i<=k;i++) 21 { 22 for(int j=0;j<=u;j++) 23 { 24 for(int t=0;t<2;t++) 25 { 26 if(t==0) 27 { 28 dp[i][j][t]=dp[i-1][j][0]+dp[i-1][j][1]; 29 } 30 else 31 { 32 dp[i][j][t]=dp[i-1][j][0]+dp[i-1][j-1][1]; 33 } 34 } 35 } 36 } 37 cout<<p<<" "<<dp[k][u][0]+dp[k][u][1]<<endl; 38 } 39 }
][0]+dp[i-1][j-1][1];
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