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 2 3 Sample Output 1 2 4 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; }
