hdu 1297
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Children’s Queue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 959 Accepted Submission(s): 534Problem 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?
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
123
Sample Output
124
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;
}
#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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