n=1 时y=sin(1)*x1 n=2 y=sin(1/2)*x1+sin(2/2)*x2 n=3 y=sin(1/3)*x1+sin(2/3)*x2+sin(3/3)*x3; n=4 y=sin(1/4)*x1+sin(2/4)*x2+sin(3/4)*x3+sin(4/4)*x4; . . . n=n y=sin(1/n)*x1+sin(2/n)*x2+.....sin(n/n)*xn; x1--xn是收盘

本帖最后由 kanhost 于 2014-4-4 17:46 编辑 int i,iLimit=1000,n; double y,temp; for(n = 1; n<= iLimit; n++) { temp=0; for(i = n-1; i >= 1; i--) { temp=temp+sin(i/n)*xi; } y=temp + sin(n/n)*xn; }

我帮你写了一个模板，你试试看！