Children’s Queue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14919 Accepted Submission(s): 4956
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
这题爆int了，表示对大数有点小阴影啊，多练习吧。加油
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <stack>
#include <vector>
#include <algorithm>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(a,b) memset(a,b,sizeof(a))
#define PI acos(-1.0)
#define LL long long
int a[1001][110] = {0};
void add(int n){
int k = 0, j;
for(j = 1; j < 101; j++){
k += a[n-1][j] + a[n-2][j] + a[n-4][j];
a[n][j] = k000;
k = k/10000;
}
/*while(k){
a[n][j++] = k % 10000;
k = k/10000;
}*/
}
int main(void){
//freopen("题.txt", "r", stdin);
a[1][1] = 1;
a[2][1] = 2;
a[3][1] = 4;
a[4][1] = 7;
int n, i;
for(i = 5; i <= 1000; i++){
add(i);
}
while(scanf("%d", &n) != EOF){
for(i = 100; i > 0; i--){
if(a[n][i] != 0) break;
}
printf("%d", a[n][i]);
for(i = i - 1; i > 0; i--){
printf("d", a[n][i]);
}
printf("\n");
}
return 0;
}
