Bnu4286--adjacent Bit Counts —————— "DP"

Adjacent Bit countstime limit:1000msmemory limit:65536kb64-bit integer IO format:%lld Java class name: Main Prev Submit Status Statistics discuss nextfor a string of NBits x 1 ,  x 2 , x 3 ,. .., x N, B the adjacent bit Countof the string ( ADJBC (x)) is given by x 1 *x 2 + x 2 *x 3 + x 3 *x 4 + ... + x N -1 *x NWhich counts the number of times a 1 bit is adjacent to another 1 bit. For EXAMPLE:ADJBC (011101101) = 3AdjBC (111101101) = 4AdjBC (010101010) = 0Write A program which takes as input integers Nand kand returns the number of bit strings xOf NBits (out of 2 ?) that satisfy ADJBC ( x) = k. For example, for 5 bit strings, there is 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 the data sets th at follow. Each data set is a contains the data set number, followed by a space, and followed by a decimal integer givin G 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) won't be greater than and the parameters N and K 'll be Chos En so, the result would fit in a signed 32-bit Integer.

Output

For each data set there are one line of output. It contains the data set number followed by a once space, followed by the number of n-bit strings with adjacent Bit count equal to K.

Sample Input

101 5 22 20 83 30 174 40 245 50 376 60 527 70 598 80 739 90 8410 100 90

Sample Output

1 62 634263 18612254 1682125015 448747646 1609167 229373088 991679 1547610 23076518


Problem-Solving ideas: Use Dp[i][j][k] to define the state. I means the current number is the I bit, J is the J value, K for the end is 0 or 1. Because when the end is 0 to the new one is not required, that is, J value does not change, so when dp[i][j][0] the transfer equation is dp[i][j][0]=dp[i-1][j][0]+dp[i-1] [j] [1]. The value of j is affected if the value of the new plus bit is 1 when the end is 1. So when Dp[i][j][1] dp[i][j][1]=dp[i-1][j-1][1]+dp[i-1][j][0].

#include <bits/stdc++.h>using namespace Std;int dp[500][500][2];void dp (int n) {    memset (dp,0,sizeof (DP));    Dp[1][0][0]=1;    Dp[1][0][1]=1;    for (int. i=2;i<n;i++) {for        (int j=0;j<i;j++) {            dp[i][j][0]=dp[i-1][j][0]+dp[i-1][j][1];            if (j) {                dp[i][j][1]=dp[i-1][j][0]+dp[i-1][j-1][1];            } else{                dp[i][j][1]=dp[i-1][j][0];}}}    int main () {    int n;    scanf ("%d", &n);    DP ();    while (n--) {        int t,ta,tb;        scanf ("%d%d%d", &T,&TA,&TB);        cout<<t<< "" <<dp[ta][tb][0]+dp[ta][tb][1]<<endl;    }    return 0;}

　　

Bnu4286--adjacent Bit Counts —————— "DP"