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//求解拉格朗日C语言程序,求

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//求解拉格朗日C语言程序,求大神指导哪错了
拉格朗日.cpp : Defines the entry point for the console application.//#include "stdafx.h"#include "math.h"#include "stdio.h"//构造求基函数的函数void lagelangri(double x,int n){double x1[6]={0.40,0.55,0.65,0.80,0.90,1.05}; double y[6]={0.41075,0.57815,0.69675,0.88811,1.02652,1.25386};int i,k; double L=0,p=0;for(i=0;i<=n;i++){ double sum=1;for(k=0;k<=n;k++)//求分母{ if(i!=k) {sum*=(x-x1[k])/(x1[i]-x1[k]);}} L+=sum*y[i];}printf("最终的运算结果为:%d\n",L);}void main(){ double x=0.596;int n=5;double x1[6]={0.40,0.55,0.65,0.80,0.90,1.05}; double y[6]={0.41075,0.57815,0.69675,0.88811,1.02652,1.25386}; lagelangri(x,n); }


来自Android客户端1楼2017-05-25 11:11回复
    //求解拉格朗日C语言程序,求大神指导哪错了
    拉格朗日.cpp : Defines the entry point for the console application.//
    #include "stdafx.h"
    # include "math.h"
    #include "stdio.h"//构造求基函数的函数void lagelangri(double x,int n){double x1[6]={0.40,0.55,0.65,0.80,0.90,1.05};
    double y[6]={0.41075,0.57815,0.69675,0.88811,1.02652,1.25386};
    int i,k;
    double L=0,p=0;
    for(i=0;i<=n;i++)
    { double sum=1;
    for(k=0;k<=n;k++)//求分母{ if(i!=k)
    {sum*=(x-x1[k])/(x1[i]-x1[k]);}}
    L+=sum*y[i];}
    printf("最终的运算结果为:%d\n",L);}
    void main(){ double x=0.596;int n=5;double x1[6]={0.40,0.55,0.65,0.80,0.90,1.05}; double y[6]={0.41075,0.57815,0.69675,0.88811,1.02652,1.25386}; lagelangri(x,n); }


    来自Android客户端2楼2017-05-25 11:15
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      2025-07-20 01:06:54
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      求大神指导,急急急,江湖救急


      来自Android客户端3楼2017-05-25 11:16
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