Children’s Queue大数加法及规律

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

123

 

Sample Output

124

 

Author

SmallBeer (CML)

 

Source

杭电ACM集训队训练赛（VIII）

 

Recommend

lcy

 

先写出前面几个来，可以发现规律f[i]=f[i-1]+f[i-2]+f[i-4];注意当1000的时候会爆long long所以要用大数！！！

代码：

#include<stdio.h>#include<string.h>char a[1005][1001];void add(int x,int y){//计算两个数相加，将各个位置的数转化为int型进行计算，然后判断是否大于10，大于10则将z=1，否则赋为0. int len_a,len_b; len_a=strlen(a[x]); len_b=strlen(a[y]); int z=0; int tx,ty,t,i=0; while(i<len_a||z==1){//如果最后z=1则继续循环将1加到末尾  tx=ty=t=0;  if(i<len_a)   tx=a[x][i]-'0';//如果i>=len_a则tx为之前赋的0值，对计算无影响了。  if(i<len_b)   ty=a[y][i]-'0';//同上。  t=tx+ty+z;//计算两个数同一位数的值，如果上两个数的值大于10则加上z的值否则为0，对计算无影响  if(t>=10){//判断此时的值是否大于10，如果大于10则t-10&&z=1   t=t-10;   z=1;  }  else//如果此时t不大于10则赋0。   z=0;  a[x][i]=t+'0';  i++; }}int main(){ int n,i; a[1][0]='1'; a[2][0]='2'; a[3][0]='4'; a[4][0]='7'; for(i=5;i<=1000;i++){  strcpy(a[i],a[i-1]);  add(i,i-2);//计算a[i-1]+a[i-2].  add(i,i-4);//计算a[i-1]+a[i-2]+a[i-4]. } while(scanf("%d",&n)!=EOF){  for(i=strlen(a[n])-1;i>=0;i--)//注意将位置颠倒过来，i=0时是各位，输出时应是最高位。   printf("%c",a[n][i]);  printf("\n"); } return 0;}