Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5687 Accepted Submission(s): 1764
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
【大数的输出花费了好久时间。】
/*题目大意：n个孩子站成一排，要求至少两个女孩站在一起。（男孩、女孩足够多），求有多少种站法。假设n-1个孩子已经站好了，现在第n个孩子过去。如果第n个孩子是男孩：则有f(n-1)种站法如果第n个孩子是女孩，那么第n-1个也必须是女孩： 如果前n-2个孩子是合法的，那么就有f(n-2)种站法 如果前n-2个孩子是不合法的，但是加上最后两个女孩就是合法的了，那么有f(n-4)种站法。n f0 11 12 23 44 7-----n可能为1000哦~所以要用大数。*/#include <stdio.h>#include <string.h>#define N 100#define M 100000long a[1001][N];void Add(long *a,long *b,long *c)//c = a + b{int i;long carry; carry = 0L;for (i=N-1;i>=0;i--) { c[i]=a[i]+b[i]+carry; carry = 0L;if (c[i]>=M) { c[i]-=M; carry=1L; } }}void output(long *a){int i=0;while(a[i] == 0L) i++; printf("%ld",a[i++]);//注意大数的输出格式。 while (i!=N) {//注意a[i]小于M-1的位数的情况 printf("ld",a[i++]); } printf("\n");}int main(){int n,i;long temp[N]; memset(a,0L,sizeof(a)); a[0][N-1]=1; a[1][N-1]=1; a[2][N-1]=2; a[3][N-1]=4;for (i=4;i<=1000;i++) {if (i==50) { i=50; } memset(temp,0L,sizeof(temp)); Add(a[i-1],a[i-2],temp); Add(temp,a[i-4],a[i]); }while (scanf("%d",&n)!=EOF) { output(a[n]); }return 0;}
转载于:https://www.cnblogs.com/CheeseZH/archive/2012/04/05/2433422.html
相关资源：数据结构—成绩单生成器