hdu 3284 Adjacent Bit Counts【dp】
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Adjacent Bit Counts
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 390 Accepted Submission(s): 320
Problem 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 2n) 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
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 2n) 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
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
题目大意:输入num,n,k。输出 num,方案数。方案数表示长度为n的串权值恰好为k的所有方案数。
思路:
简单dp求解。
我们设dp【i】【j】【k】表示长度为i最后一个字母为j的权值为k的串的方案数。
那么不难推出:
dp【i】【0】【k】=dp【i-1】【0】【k】+dp【i-1】【1】【k】;//最后一个字母如果为0的话,那么方案数应该来自于长度i-1的最后一个字母任意的方案数加和
dp【i】【1】【k】=dp【i-1】【0】【k】+dp【i-1】【1】【k-1】;//最后一个字母如果为1的话,那么方案数应该来自于长度为i-1最后一个字母为0的方案数和长度为i-1最后一个字母为1的权值为k-1的方案数的加和。
注意处理权值为0的时候的方案数。
AC代码:
#include<stdio.h>#include<string.h>using namespace std;int dp[105][2][105];//i j quanzhi kint main(){ memset(dp,0,sizeof(dp)); dp[1][1][0]=dp[1][0][0]=1; for(int i=2;i<=105;i++) { dp[i][0][0]=dp[i-1][0][0]+dp[i-1][1][0]; dp[i][1][0]=dp[i-1][0][0]; for(int j=1;j<i;j++) { dp[i][0][j]=dp[i-1][0][j]+dp[i-1][1][j]; dp[i][1][j]=dp[i-1][1][j-1]+dp[i-1][0][j]; } } int t; scanf("%d",&t); while(t--) { int num,n,k; scanf("%d%d%d",&num,&n,&k); printf("%d %d\n",num,dp[n][0][k]+dp[n][1][k]); }}
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