http://acm.hdu.edu.cn/showproblem.php?pid=5366
Problem Description
![](../../data/images/C613-1001-1.jpg)
ZJiaQ want to become a strong man, so he decided to play the mook jong。ZJiaQ want to put some mook jongs in his backyard. His backyard consist of n bricks that is 1*1,so it is 1*n。ZJiaQ want to put a mook jong in a brick. because of the hands of the mook jong,
the distance of two mook jongs should be equal or more than 2 bricks. Now ZJiaQ want to know how many ways can ZJiaQ put mook jongs legally(at least one mook jong).
Input
There ar multiply cases. For each case, there is a single integer n( 1 < = n < = 60)
Output
Print the ways in a single line for each case.
Sample Input
1 2 3 4 5 6
Sample Output
1 2 3 5 8 12
/** hdu 5366 排列组合 题目大意:ZJiaQ为了强身健体,决定通过木人桩练习武术。ZJiaQ希望把木人桩摆在自家的那个由1*1的地砖铺成的1*n的院子里。 由于ZJiaQ是个强迫症,所以他要把一个木人桩正好摆在一个地砖上,由于木人桩手比较长,所以两个木人桩之间地砖 必须大于等于两个,现在ZJiaQ想知道在至少摆放一个木人桩的情况下,有多少种摆法 解题思路:标准题解说是找规律,我智商太低了,只会排列组合,不过也过了。说一下我想法:由于相邻的两个木人之间距离大于1,那么假设现在要插入x个 木人,那么我们把x-1的木人每个占3个的地板看作一个,那么地板总数就为ans=n-(x-1)*2个地板,答案就是C[ans][x],枚举一下x就可以了。 对于c数组到60会爆掉long long 不过我们注意到x最大不过21那么就没有问题了,前面的是不会爆的 */ #include <stdio.h> #include <string.h> #include <algorithm> #include <iostream> using namespace std; typedef long long LL; int n; LL a[60][60]; void judge() { a[0][0]=1; a[1][1]=a[1][0]=1; for(int i=2;i<=60;i++) { a[i][0]=1,a[i][i]=1; for(int j=1;j<i;j++) { a[i][j]=a[i-1][j]+a[i-1][j-1]; } }/* for(int i=1;i<=25;i++) { for(int j=0;j<=i;j++) { printf("%d ",a[i][j]); } printf("\n"); }*/ } int main() { judge(); while(~scanf("%d",&n)) { LL sum=0; for(int i=1;;i++) { if((i-1)*3+1>n)break; int x=n-2*(i-1); sum+=a[x][i]; } printf("%I64d\n",sum); } return 0; }
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时间: 2024-10-09 02:51:46