HDOJ 题目1297Children’s Queue（递推，大数）

发布时间：2020-12-14 03:03:01 所属栏目：大数据来源：网络整理

导读：?? Children’s Queue Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10644????Accepted Submission(s): 3431 Problem Description There are many students in PHT School. One day,the head

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10644????Accepted Submission(s): 3431

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

Sample Output

Author

SmallBeer (CML)

Source

杭电ACM集训队训练赛（VIII）

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递推公式；a[i]=a[i-1]+a[i-2]+a[i-4];

ac代码

#include<stdio.h>
#include<string.h>
char a[1010][1010];
char *add(char a[],char b[],char tc[])
{
	int c[10000];
	int i,j,len1,len2,k=0,max;
	memset(c,sizeof(c));
	len1=strlen(a);
	len2=strlen(b);
	i=len1-1;
	j=len2-1;
	memset(c,sizeof(c));
	while(i>=0||j>=0)
	{
		if(i>=0&&j<0)
			c[k]+=a[i]-'0';
		else
			if(i<0&&j>=0)
				c[k]+=b[j]-'0';
			else
				c[k]+=b[j]-'0'+a[i]-'0';
			k++;
			c[k]=c[k-1]/10;
			c[k-1]%=10;
			i--;
			j--;
			if(c[k])
				max=k;
			else
				max=k-1;
	}
	j=0;
	for(i=max;i>=0;i--)
	{
		tc[j++]=c[i]+'0';
	}
	tc[j]='';
	return tc;
}
void fun()
{
	int i;
	//memset(a,sizeof(a));
	strcpy(a[0],"1");
	strcpy(a[1],"1");
	strcpy(a[2],"2");
	strcpy(a[3],"4");
	for(i=4;i<1010;i++)
	{
		add(a[i-1],a[i-2],a[i]);
		add(a[i-4],a[i],a[i]);
	}
}
int main()
{
	int n;
	fun();
	while(scanf("%d",&n)!=EOF)
	{
		printf("%sn",a[n]);
	}
}

（编辑：李大同）

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