Adjacent Bit Counts
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 599 |
|
Accepted: 502 |
Description
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?) 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
Source
-
#include <stdio.h>
#include <math.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <queue>
#define maxn 105
using namespace std;
int main()
{
int t,n,m;
int dp[maxn][maxn][2];
dp[1][0][0] = dp[1][0][1] = 1;
for(int i=2;i<maxn;i++){
dp[i][0][0] = dp[i-1][0][1] + dp[i-1][0][0];
dp[i][0][1] = dp[i-1][0][0];
}
for(int i=2;i<maxn;i++){
for(int j=1;j<maxn;j++){
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];
}
}
int T;
cin >> T;
while(T--){
cin >> t >> n >> m;
cout << t << " " << dp[n][m][0]+dp[n][m][1] << endl;
}
return 0;
}
POJ 3786 dp-递推 Adjacent Bit Counts *
原文：http://www.cnblogs.com/l609929321/p/7238061.html
