首页 > ACM题库 > HDU-杭电 > HDU 3284-Adjacent Bit Counts-动态规划-[解题报告]HOJ
2014
03-13

HDU 3284-Adjacent Bit Counts-动态规划-[解题报告]HOJ

Adjacent Bit Counts

问题描述 :

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

输入:

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.

输出:

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.

样例输入:

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

样例输出:

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

题目链接:Click here~~

题意:

一种只有0、1两种元素的串,每个串有一个权值 x1*x2 + x2*x3 + x3*x4 + … + xn-1*xn。给出你某个串的长度 n,求出其权值为 k 时的方案种数。

解题思路:

这么水的DP都不会,哎。好像见过几道关于01串的dp问题了,下次遇到可以往这方面想想。

分析:对于长度为 i 的串,假设它的权值为 j,则长度为 i+1 的串的权值只可能为 j 或 j+1,且仅与末位元素和新添加元素有关。

令 dp[i][j][k] 表示长度为 i 的串、权值为 j 、末位为 k (0 or 1) 的方案种数。

状态转移方程为 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]。

确实会溢出 int && long long ,没考虑, 懒得写大数了,原谅我吧。。。

#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;
}

参考:http://blog.csdn.net/dgq8211/article/details/8041473


  1. #include <cstdio>

    int main() {
    //answer must be odd
    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;
    }