顺推法即由边界条件出发，通过递推关系式推出后项值，再由后项值按递推关系式推出再后项值。。。依次递推，直至从问题初始陈述向前推进到这个问题的解为止。

实例

代码

#include<iostream>
#include<stdlib.h>
using namespace std;

const int maxN = 60 ;
int N = 4 ,d = 2 , m = 3;
float a1 = 2 ,an = 5;

void input()
{
    cout<<"input data N,d,a1,an,m"<<endl;
    cin>>N>>d>>a1>>an>>m;
    if( N > maxN )
    {
        cout<<"input error"<<endl;
        return ;
    }

    if( N > m )
    {
        cout<<" m is too large "<<endl;
        return ;
    }

    if( m < 0 )
    {
        cout<<"m must be a positive data."<<endl;
        return;
    }

}

//计算am与a1之间的关系s
//a[i] = s[m][0]*a[2] + s[i][1]*d + s[i][2]*a1
//a2 = ( an - s[N-1][1] * d - s[N-1][2] * a1 ) / s[N-1][0]
void calculate_s(float  **s ,float *list)
{

    s[0][0] = 0;
    s[0][1] = 0;
        s[0][2] = 1;

    s[1][0] = 1;
    s[1][1] = 0;
        s[1][2] = 0;
    for(int i = 2 ; i < N ; i++ )
    {
        s[i][0] = s[i-2][0] - 2 * s[i-1][0];
        s[i][1] = s[i-2][1] - 2 * s[i-1][1] + 2;
        s[i][2] = s[i-2][2] - 2 * s[i-1][2];
    }
    list[0] = a1;
    list[1] = ( an - s[N-1][1] * d - s[N-1][2] * a1 ) / s[N-1][0] ;
    for(int j = 2 ; j < N ; j++ )
    {
        list[j] = list[j-2] - 2*list[j-1] +2 * d;
    }

}

//输出最后的结果
void out(float  **s ,float *list)
{
    int i;

    for(i = 0 ;i < N ; i++ )
    {
        cout<<s[i][0]<<" "<<s[i][1]<<" "<<s[i][2]<<" "<<endl;
    }

    cout<<"data........"<<endl;
    for(i = 0 ;i < N ; i++ )
    {
        cout<<list[i]<<" ";
    }

    cout<<endl;
}


int main()
{
    //input();

    //s[i][0]表示p s[i][1]表示q s[i][2]表示r
    float **s = (float **)malloc(sizeof(float*) * N);
    for(int ii = 0 ; ii < N ; ii++ )
        s[ii] = (float *)malloc(sizeof(float) * 3); 
    float *list = (float *)malloc( sizeof(float) * N ); 

    calculate_s(s , list );

    out( s , list );

    cout<<"\nthe result of "<<m<<"  data is:";
    cout<<list[m-1]<<endl;

    return 0;
}