POJ 3786 dp-遞推 Adjacent Bit Counts *
阿新 • • 發佈：2017-07-26
num new red rip int chosen star program http Adjacent Bit Counts
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 2?) 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
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 599 | Accepted: 502 |
Description
For a string of n bits x1, x2, x3, …, xn, the adjacent bit count of the string (AdjBC(x)) is given byx1*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 2?) 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
Output
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
Source
Greater New York Regional 2009- 由0,1組成的長度為n的數列x1,x2,x3,x4.....xn,定義一個操作為
- AdjBC(x) = x1*x2 + x2*x3 + x3*x4 + … + xn-1*xn 。
- 輸入兩個數n和m,求長度為n的數列,經過上述操作,最後結果為m共有多少種
- 思路 : DP，遞推
- d[i][j][0]：表示前i項組成和為j且第i項為0共有多少種
- d[i][j][1]：表示前i項組成和為j且第i項為1共有多少種
- 狀態轉移方程：
- d[i][j][1] = d[i-1][j][0] + d[i-1][j-1][1];
- d[i][j][0] = d[i-1][j][1] + d[i-1][j][0]
#include <stdio.h> #include <math.h> #include <iostream> #include <string.h> #include <algorithm> #include <queue> #define maxn 105 using namespace std; int main() { int t,n,m; int dp[maxn][maxn][2]; dp[1][0][0] = dp[1][0][1] = 1; for(int i=2;i<maxn;i++){ dp[i][0][0] = dp[i-1][0][1] + dp[i-1][0][0]; dp[i][0][1] = dp[i-1][0][0]; } for(int i=2;i<maxn;i++){ for(int j=1;j<maxn;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]; } } int T; cin >> T; while(T--){ cin >> t >> n >> m; cout << t << " " << dp[n][m][0]+dp[n][m][1] << endl; } return 0; }
POJ 3786 dp-遞推 Adjacent Bit Counts *