HDU 1297 Children’s Queue【大数加法 + 递推】

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

分析： 找到递推式                                  // dp[n]=dp[n-1]*2-dp[n-2]+dp[n-3]

          但为了便于大数加法 把递推式改为    // dp[n]=dp[n-1]+dp[n-2]+dp[n-4]

code：

// dp[n]=dp[n-1]<<1-dp[n-2]+dp[n-3]
// dp[n]=dp[n-1]+dp[n-2]+dp[n-4]
#include<stdio.h>  
#include<string.h>
int dp[1001][302];
void pl(int s1[],int s2[],int s3[],int s4[])
{
    int i,c;
    for(i=0;i<=300;i++)
        s4[i]+=s1[i]+s2[i]+s3[i];
    for(i=c=0;i<=300;i++)
    {
        s4[i]+=c;
        c=s4[i]/10;
        s4[i]=s4[i]%10;
    }
}

int main()
{
    int i,j,n;
    memset(dp,0,sizeof(dp));
    dp[1][0]=1;  dp[2][0]=2;  dp[3][0]=4;  dp[4][0]=7;
    for(i=5;i<=1000;i++)
        pl(dp[i-1],dp[i-2],dp[i-4],dp[i]);
    while(scanf("%d",&n)!=EOF)
    {
        for(i=300;i>=0;i--)
            if(dp[n][i])
                break;
        for(j=i;j>=0;j--)
            printf("%d",dp[n][j]);
        printf("\n");
    }
    return 0;
}