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开始输入(xi,yi)及 x i =0,1,2,.,ny=0,k=0t=1t=t(x-xi)/(xk-xi) i=0,1,2,.,k-1, k+1,.,ny=y+yktk=nk=k+1输出y结束FT编程思想(算法 )注意:内、外两个 循环的,内循环i, 和外循环kLagrange插值步骤:1、输入数组x,y的值,令Ln(x0)=02、对i=0,1n计算#include #include #define N 4float Lagrange(float xN,float fN,float xx) float yy=0,t; int i,j; for(i=0;iN;i+) printf(“x%d=%f “,i,xi); for(i=0;iN;i+) t=1.0;for(j=0;jN;j+)if(j!=i)t=t*(xx-xj)/(xi-xj);yy=yy+t*fi; return yy; main() float xN=0.46,0.47,0.48,0.49,fN=0.4846555,0.4937452,0.5047498,0.5 116683,xx=0.472,yy;yy=Lagrange(x,f,xx);printf(“yy(%lf)=%.7lf“,xx,yy);