hdu 1297
Children’s Queue |
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
Total Submission(s): 959 Accepted Submission(s): 534 |
Problem Description
There 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 |
Sample Output
1 大数处理加dP,,dp的话f[n]=f[n-1]+f[n-2]+f[n-4]
假设当前要加入第n个学生,如果他是男生,则加入对前面的序列都不会构成影响,
即为f[n-1],假如他是女生,为保证合法,她前面的那个也就是第n-1个也必须是
女生,接下来分两种情况讨论,如果第n-2个及以前的都合法,那么显然加入最后
这两个女生也是合法的额,但是如果不合法呢?很明显,如果不合法,问题应该出在
尾部,也就是原来f[n-2]不合法,但是加入最后两个女生后也就变得合法了(这句话
很关键),那么其只能为f[n-4]+男+女,所以即可得到公式
其实这个讨论的关键在于确定最后两个为女生才能保证合法
#include<iostream>
#include<cstdio> #include<cstring> using namespace std; long long dp[1001][1001]; int main() { int n; memset(dp,0,sizeof(dp)); dp[1][0]=1;dp[1][1]=1; dp[2][0]=1;dp[2][1]=2; dp[3][0]=1;dp[3][1]=4; dp[4][0]=1;dp[4][1]=7; for(int i=5;i<=1000;i++) { int j,t=dp[i-1][0]; for(j=1;j<=t;j++) { dp[i][j]+=dp[i-1][j]+dp[i-2][j]+dp[i-4][j]; if(dp[i][j]>=10) { dp[i][j+1]+=dp[i][j]/10; dp[i][j]%=10; } } while(dp[i][j]>=10) { dp[i][j+1]+=dp[i][j]/10; dp[i][j]%=10; j++; } j=1000; while(!dp[i][j]) j--; dp[i][0]=j; } while(cin>>n) { for(int i=dp[n][0];i>=1;i--) cout<<dp[n][i]; cout<<endl; } return 0; } |
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