递推（大数处理）

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9545    Accepted Submission(s): 3049


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

递推式去杭电ppt看，fn=fn-1+fn-2+fn-4.这个求出来fn的值会非常大，需要用数组存。

　　

#include<stdio.h>
#include<iostream>
#include<string>
#define MAXM 1001
#define MAXN 502
using namespace std;
int a[MAXM][MAXN];
int main()
{
    int n, t, p, i, j;
    memset(a, 0, sizeof(a));
    a[1][0] = 1; a[2][0] = 2; a[3][0] = 4; a[4][0] = 7;
    for (i = 5; i<MAXM; i++){
        for (j = 0; j<MAXN; j++){
            a[i][j] += a[i - 1][j] + a[i - 2][j] + a[i - 4][j];
            if (a[i][j] >= 10){
                a[i][j + 1] += a[i][j] / 10;
                a[i][j] %= 10;
            }
        }
    }
    while (scanf("%d", &n) != -1){
        for (i = 500; a[n][i] == 0; i--); 
        for (; i>-1;i--)
        {
            printf("%d", a[n][i]);
        }
        printf("
");
    }
    return 0;
}