HDUOJ 1297 - Children’s Queue(递推求解 + string大数相加)
2017-11-23 16:29
429 查看
原题:
Problem DescriptionThere are many students in PHT School. One day, the headmaster whose name is PigHeader wanted all students stand in a line. He prescribed that girl can not be in single. In other words, either no girl in the queue or more than one girl stands side by side. The case n=4 (n is the number of children) is like
FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM
Here F stands for a girl and M stands for a boy. The total number of queue satisfied the headmaster’s needs is 7. Can you make a program to find the total number of queue with n children?
Input
There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer n means the number of children (1<=n<=1000)
Output
For each test case, there is only one integer means the number of queue satisfied the headmaster’s needs.
Sample Input
1
2
3
Sample Output
1
2
4
解题思路:
按照最后一个人的性别分析,他要么是男,要么是女,所以可以分两大类讨论:如果n个人的合法队列的最后一个人是男,则对前面n-1个人的队列没有任何限制,他只要站在最后即可,所以,这种情况一共有F(n-1);
如果n个人的合法队列的最后一个人是女,则要求队列的第n-1个人务必也是女生,这就是说,限定了最后两个人必须都是女生,这又可以分两种情况:
如果队列的前n-2个人是合法的队列,则显然后面再加两个女生,也一定是合法的,这种情况有F(n-2);
但是,难点在于,即使前面n-2个人不是合法的队列,加上两个女生也有可能是合法的,当然,这种长度为n-2的不合法队列,不合法的地方必须是尾巴,就是说,这里说的长度是n-2的不合法串的形式必须是“F(n-4)+男+女”,这种情况一共有F(n-4).
所以,通过以上的分析,可以得到递推的通项公式:
F(n)=F(n-1)+F(n-2)+F(n-4) (n>3)
然后就是对n<=3 的一些特殊情况的处理了
显然:F(0)=1 (没有人也是合法的,这个可以特殊处理,就像0的阶乘定义为1一样)
F(1)=1
F(3)=4
代码:
递推求解–找递推公式
#include<stdio.h> #include <string> #include <iostream> using namespace std; string dp[1001] = { "1","1","2","4" }; string BigNumAdd(string a, string b) { int A_Len = a.length() - 1, B_Len = b.length() - 1, i, temp, CarryBit = 0; //前面补0,弄成长度相同 if (A_Len<B_Len) { for (i = 1;i <= B_Len - A_Len;i++) a = "0" + a; } else { for (i = 1;i <= A_Len - B_Len;i++) b = "0" + b; } //------ for (i = A_Len; i >= 0; i--) { temp = a[i] - '0' + (b[i] - '0') + CarryBit; CarryBit = temp / 10;//进位 temp %= 10; a[i] = char(temp + '0'); } if (CarryBit != 0)a = char(CarryBit + '0') + a; return a; } int main() { int n, i; for (i = 4; i <= 1000; i++) dp[i] = BigNumAdd(BigNumAdd(dp[i - 1], dp[i - 2]), dp[i - 4]); while (scanf("%d", &n) != EOF) cout << dp << endl; }
相关文章推荐
- 递推—杭电1297 Children’s Queue
- HDU 1297 Children’s Queue【大数加法 + 递推】
- 【大数递推】HDU 1297——Children’s Queue
- HDU 1297 Children’s Queue(递推)
- hdu 1297 Children’s Queue(递推和高精)
- hdu 1297 Children’s Queue (大数加法+递推)
- 杭电ACM1297——Children’s Queue~~大数相加的应用
- HDU-1297 Children’s Queue(递推)(高精度)
- HDOJ/HDU 1297 Children’s Queue(推导~大数)
- hdu 1297 Children’s Queue 递推 大数
- Children’s Queue(递推 + JAVA大数)
- HDOJ/HDU 1297 Children’s Queue(推导~大数)
- Children’s Queue HDU 1297 递推+大数
- 【HDU】1297 - Children’s Queue(BigDecimal & 递推 & 思维)
- 杭电acm 1297 Children's Queue.大数加递归
- hdu 1297 Children’s Queue(大数处理)
- HDU 1297 Children’s Queue 递推
- Hduoj1047【大数相加】
- ACM--学生排队--HDOJ 1297--Children’s Queue--大数
- 杭电1297 Children’s Queue