2014
03-13

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:

Write a program which takes as input integers n and k and returns the number of bit strings x of n bits (out of 2n) that satisfy AdjBC(x) = k. For example, for 5 bit strings, there are 6 ways of getting

11100, 01110, 00111, 10111, 11101, 11011

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.

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.

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

1 6
2 63426
3 1861225
4 168212501
5 44874764
6 160916
7 22937308
8 99167
9 15476
10 23076518

#include <stdio.h>

int dp[102][102][2];

int main()
{
int z,ca,n,k;
scanf("%d",&z);
dp[1][0][0] = dp[1][0][1] = 1;
for(int i=2;i<=100;i++)
{
dp[i][0][0] = dp[i-1][0][0] + dp[i-1][0][1];
dp[i][0][1] = dp[i-1][0][0];
for(int j=1;j<i;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];
}
}
while(z--)
{
scanf("%d%d%d",&ca,&n,&k);
printf("%d %d\n",ca,dp[n][k][0]+dp[n][k][1]);
}
return 0;
}

1. #include <cstdio>

int main() {
int n, u, d;
while(scanf("%d%d%d",&n,&u,&d)==3 && n>0) {
if(n<=u) { puts("1"); continue; }
n-=u; u-=d; n+=u-1; n/=u;
n<<=1, ++n;
printf("%dn",n);
}
return 0;
}