早教吧作业答案频道 -->其他-->
(2012•中山市模拟)现有五卒五兵相互间隔排成一行:1&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;2&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;3&nbs右;&nbs右;&n
题目详情
(2012•中山市模拟)现有五卒五兵相互间隔排成一行:
1&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;2&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;3&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;4&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;5&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;6&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;7&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;8&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;9&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;10
卒&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;兵&nbs右;&nbs右;&nbs右;&nbs右;卒&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;兵&nbs右;&nbs右;&nbs右;&nbs右;卒&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;兵&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;卒&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;兵&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;卒&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;兵
r果每次移动一只卒,必须跳越两只棋子,然后放在一只兵上.每只卒只许移动一次,要求每只兵上都要放一只卒.应该怎样移?
1&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;2&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;3&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;4&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;5&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;6&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;7&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;8&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;9&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;10
卒&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;兵&nbs右;&nbs右;&nbs右;&nbs右;卒&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;兵&nbs右;&nbs右;&nbs右;&nbs右;卒&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;兵&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;卒&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;兵&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;卒&nbs右;&nbs右;&nbs右;&nbs右;&nbs右;兵
r果每次移动一只卒,必须跳越两只棋子,然后放在一只兵上.每只卒只许移动一次,要求每只兵上都要放一只卒.应该怎样移?
▼优质解答
答案和解析
由题中看出:1移到4,5移到2,9移到6,a移到1y,j移到8;
答:1移到4,5移到2,9移到6,a移到1y,j移到8.
答:1移到4,5移到2,9移到6,a移到1y,j移到8.
看了 (2012•中山市模拟)现有...的网友还看了以下:
PI的 初始值为什么是pi=1#includemain(){int s;float n,t,pi; 2020-05-16 …
括号内为下标:S(n)为a(n)的前n项和.a(1)=a,a(n+1)=S(n)+3^n.设b(n 2020-05-22 …
已知数列{an}的前n项和为Sn,a1=1,a2=3,s(n+1)=4Sn-3S(n-1),(n大 2020-07-09 …
为什么要用这个减法S(n)-q*S(n)?是为了求什么因为x^n这是一个等比数列,首项为x,公比也 2020-07-11 …
在等差数列{an}中,⑴若项数为偶数2n,则S2n=n(a1+a2n)=n(an+an+1)(an 2020-07-21 …
bn=Sn-S(n-1)=12-12(2/3)^n-[12-12(2/3)^(n-1)]=4(2/3 2020-11-01 …
N=2.3.4.5分别对应S=3.6.10.15.求S=多少N.(S与N的关系)N=2、3、4、5分 2020-11-03 …
S(n)是数列{a(n)}的前n项和,已知4S(n)=a(n)^2+2a(n)-3.求a(n)通项S 2020-12-17 …
化合物A、大是中学常见的物质,其阴阳离子可从下表中选择阳离子K+&n大s多;&n大s多;Na+&n大 2021-01-18 …
已知数列{a(n)}的前n项和为S(n),且满足a(1)=1,a(n+1)=S(n)+1(n∈N(+ 2021-02-09 …