【C/C++】龙格库塔+亚当姆斯求解数值微分初值问题

 /*
     解数值微分初值问题：
     龙格-库塔法求前k个初值 + 亚当姆斯法
 */
 #include<bits/stdc++.h>
 using namespace std;

 double f(double x,double y){
     //y(0) = 1
     return (y - *x/y);
 }
 void getRungeResult(double *Runge_k,double x0,double y0,double h,int N){
     //求解N个初值，保存在Runge_k[1 to N]中
     double K1,K2,K3,K4;
     double x1,y1;
     for(int i = ;i<=N;i++){
         x1 = x0+h;
         K1 = f(x0,y0);
         K2 = f(x0+h/,y0+h/*K1);
         K3 = f(x0+h/,y0+h/*K2);
         K4 = f(x1,y0+K4);
         y1 = y0 + h/*(K1+*K2+*K3+K4);
         Runge_k[i] = y1;
         x0 = x1;
         y0 = y1;
     }
     return;
 }

 //亚当姆斯多步法
 void Adams(double *Runge_k,double *predict,double x0,double y0,double h,int N){
     Runge_k[] = y0;
     //（0）龙格库塔法求前4个初值
     getRungeResult(Runge_k,x0,y0,h,);
     double y1,y2,y3,dy0,dy1,dy2,dy3;
     y1 = Runge_k[];
     y2 = Runge_k[];
     y3 = Runge_k[];
     dy0 = f(x0,y0);
     dy1 = f(x0+h,y1);
     dy2 = f(x0+*h,y2);
     dy3 = f(x0+*h,y3);
     double x3 = x0+*h;
     double x4,y4,yp,dyp,dy4;
     for(int i = ;i<=N;i++){
         x4 = x3+h;
         //(1)预测
         yp = y3 + h/*(*dy3-*dy2+*dy1-*dy0);
         predict[i] = yp;//保存预测值
         //预测要用dyp
         dyp = f(x4,yp);
         //(2)校正
         y4 = y3 + h/*(*dyp + *dy3 -*dy2+dy1);
         //存起来
         Runge_k[i] = y4;
         //求下一次需要用到导
         dy4 = f(x4,y4);
         //为下一次循环做准备
         x3 = x4;
         y3 = y4;
         dy0 = dy1;
         dy1 = dy2;
         dy2 = dy3;
         dy3 = dy4;
     }
     return;
 } 

 /*假设这里保证四阶精度*/
 int main(){
     /*说明：x0,y0是初值，h是小区间长度，N是要求的个数*/
     double x0,y0,h;
     int N;
     cout<<"输入初值x0,y0,小区间h,需要的初值个数N：";
     cin>>x0>>y0>>h>>N;
     //保存Runge求的4个初始值,龙格法求3个就可以;之后也用这个保存最终的Adams结果
     double Runge_k[];
     //保存预测值，方便以后比较
     double predict[];
     memset(predict,,sizeof(predict));
     memset(Runge_k,,sizeof(Runge_k));
     Adams(Runge_k,predict,x0,y0,h,N);
     cout<<endl;
     printf("预测值：");
     for(int i = ;i<=N;i++){
         if(i<){
             printf("%.6lf ",);
         }else{
             printf("%.6lf ",predict[i]);
         }
     }
     cout<<endl;
     printf("校正值：");
     for(int i = ;i<=N;i++){
         printf("%.6lf ",Runge_k[i]);
     }

 }