# 杰拉斯的博客

## [ACM_HDU_1297]Children’s Queue

### Children’s Queue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5660 Accepted Submission(s): 1755

#### 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.

```
1
2
3
```

```
1
2
4
```

#### Source

HDU1297

a.男孩，任何n - 1的合法队列追加1个男孩必然是合法的，情况数为f[n - 1]；

b.女孩，在前n - 1的以女孩为末尾的队列后追加1位女孩也是合法的，我们可以转化为n - 2的队列中追加2位女孩；

`f[n] = f[n - 1] + f[n - 2] + f[n - 4]`

`12748494904808148294446671041721884239818005733501580815621713101333980596197474744336199742452912998225235910891798221541303838395943300189729514282623665199754795574309980870253213466656184865681666106508878970120168283707307150239748782319037`

```#include<stdio.h>
int main(){
int n;
int f[1001][101] = {0};
f[0][1] = 1;
f[1][1] = 1;
f[2][1] = 2;
f[3][1] = 4;
for(int i = 4; i < 1001; ++i){
for(int j = 1; j < 101; ++j){
f[i][j] += f[i - 1][j] + f[i - 2][j] + f[i - 4][j];	//数组的每一位相加
f[i][j + 1] += f[i][j] / 10000;	//超过4位的部分保存至数组下一位中
f[i][j] %= 10000;	//每位数组只保存其中4位
}
}
while(scanf("%d", &n) != EOF){
int k = 100;
while(!f[n][k--]);	//排除前面为空的数组
printf("%d", f[n][k + 1]);	//输出结果的前四位
for(; k > 0; --k){
printf("%04d", f[n][k]);	//输出其余的所有四位数字，若数字小于四位，则前面用0填充
}
printf("\n");
}
return 0;
}
```

#### 5 条评论 »

1. housansan

在前n - 1的以女孩为末尾的队列后追加1位女孩也是合法的，我们可以转化为n - 2的队列中追加2位女孩；

为什么要追加换成2位。不换时就是 f[n - 3]。
为什么我这么理解不行？？

1. 不换时就是f[n - 3]？怎么理解?
n - 3 的话那还需要追加3个才可以达到 n，这3个要怎么安排呢？

2. 青鸢不死

大数计算为什么定义k=100

1. 因为前面定义的数组：
int f1001;
所以从100开始往后查。

3. 不会撒谎的乌索普

没看懂。