nyoj 715 Adjacent Bit (码数+打表)_第六届河南省程序设计大赛

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于 2018-04-30 21:45:47 发布

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分类专栏：算法与数据结构

本文链接：https://blog.youkuaiyun.com/qq_36914923/article/details/80152447

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本文介绍了一种计算特定长度的二进制字符串中相邻1的数量等于给定值的方法。通过预处理技术，文章提供了一个高效的解决方案，并给出了具体的实现代码。

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Adjacent Bit Counts

时间限制：1000 ms | 内存限制：65535 KB

难度：4

描述

For a string of n bits x1,x2,x3,…,xn,the adjacent bit count of the string is given by fun(x)=x₁*x₂+x₂*x₃+x₃*x₄+…+x_n-1*x_nwhich counts the number of times a 1 bit is adjacent to another 1 bit.For example:

Fun(011101101)=3

Fun(111101101)=4

Fun(010101010)=0

Write a program which takes as input integers n and p and returns the number of bit strings x of n bits(out of 2ⁿ)that satisfy Fun(x)=p.

For example,for 5 bit strings,there are 6 ways of getting fun(x)=2:

11100,01110,00111,10111,11101,11011

输入

On the first line of the input is a single positive integer k,telling the number of test cases to follow.1≤k≤10 Each case is a single line that contains a decimal integer giving the number(n)of bits in the bit strings,followed by a single space,followed by a decimal integer(p)giving the desired adjacent bit count.1≤n,p≤100

输出

For each test case,output a line with the number of n-bit strings with adjacent bit count equal to p.

样例输入

5 2

20 8

样例输出

63426

题意：长度为n的0,1字符串按照下面的公式，组成值为p的个数1≤n,p≤100；

fun(x)=x₁*x₂+x₂*x₃+x₃*x₄+…+x_n-1*x_n

例如：

Fun(011101101)=3

Fun(111101101)=4

Fun(010101010)=0

f[i][j][0] 长度为j的串第j位是0时组成值为i的个数；f[i][j][0]=f[i][j-1][1]+f[i][j-1][1];

f[i][j][1] 长度为j的串第j位是1时组成值为i的个数；f[i][j][1]=f[i-1][j-1][1]+f[i-1][j-1][0];

初始化：f[0][1][0]=f[0][1][1]=1;

#include<stdio.h>

long long f[105][105][2];

void init()
{//打表
	f[0][1][0]=1;
	f[0][1][1]=1;
	for(int i=2;i<=100;i++)
	{
		f[0][i][0]=f[0][i-1][0]+f[0][i-1][1];
		f[0][i][1]=f[0][i-1][0];
	}
	for(int i=1;i<=100;i++)
		for(int j=2;j<=100;j++)
		{
			f[i][j][0]=f[i][j-1][1]+f[i][j-1][0];
			f[i][j][1]=f[i-1][j-1][1]+f[i][j-1][0];
		}
}
int main()
{
	init();
	int k,n,p;
	scanf("%d",&k);
	while(k--)
	{
		scanf("%d%d",&n,&p);
		printf("%lld\n",f[p][n][0]+f[p][n][1]);
	}
}