编程题

c语言实现编辑本段问题已知f(x)=x*e^x-1针对f(x)=0类型。迭代方程是：g(x)=x-f(x)/f"(x)；其中f"(x)是导数。针对x*e^x-1=0的牛顿迭代法求出迭代方程，根据牛顿的是，g(x)=x-f(x)/f"(x)针对x*e^x-1=0，是g(x)=x-(xe^x-1)/(e^x+x*e^x);代码#include#includeintmain(){doublef(doublex);doublex,y,d;x=1;y=0;//迭代值。d=0.000000001;//误差控制

Private Function NewTon(n As Double, eps As Double) As Double Dim a(100) As Double Dim x As Double Dim y As Double Dim k As Double a(0) = 0 For i = 0 To 99 x = a(i) y = x ^ 3 - 2 * x ^ 2 + 4 * x + 1 k = 3 * x ^ 2 - 4 * x + 4 a(i + 1) = x - y / k Print a(i + 1) If Abs(y) < esp Then NewTon = a(i + 1) Exit Function Else NewTon = 0 End If Next iEnd Function

通过官网查询。1、首先打开赛码编程官网。2、在赛码编程官网查询系统中输入自己的名字与比赛序号。3、然后点击查询即可知道通过没有。

matlab 的程序要不？a=zeros(1,5);b=zeros(1,5);c=zeros(1,5);d=zeros(1,5);u=zeros(1,5);x=zeros(1,5);y=zeros(1,5);a(:)=-1;b(:)=4;c(:)=-1;d(:)=[100,0,0,0,200];u(1)=c(1)/b(1);y(1)=d(1)/b(1);for n=2:4 u(n)=c(n)/(b(n)-u(n-1)*a(n));endfor t=2:5 y(t)=(d(t)-y(t-1)*a(t))/(b(t)-u(t-1)*a(t));endx(5)=y(5);for s=-2:-1:-5 x(s+6)=y(s+6)-u(s+6)*x(s+7);endx%%下一问题一样：a=zeros(1,10);b=zeros(1,10);c=zeros(1,10);d=zeros(1,10);u=zeros(1,10);x=zeros(1,10);y=zeros(1,10);a(:)=1;b(:)=-2;c(:)=1;d(:)=[-0.5,-1.5,-1.5,-1.5,-1.5,-1.5,-1.5,-1.5,-1.5,-0.5];u(1)=c(1)/b(1);y(1)=d(1)/b(1);for n=2:9 u(n)=c(n)/(b(n)-u(n-1)*a(n));endfor t=2:10 y(t)=(d(t)-y(t-1)*a(t))/(b(t)-u(t-1)*a(t));endx(10)=y(10);for s=-2:-1:-10 x(s+11)=y(s+11)-u(s+11)*x(s+12);endx

clearx=[.2 .4 .6 .8 1.0];y=[0.98 0.92 0.81 0.64 0.38];yi0=interp1(x,y,0.025,"linear")xi=0:.02:1;yi=interp1(x,y,xi,"linear");zi=interp1(x,y,xi,"spline");wi=interp1(x,y,xi,"cubic");plot(x,y,"o",xi,yi,"r+",xi,zi,"g*",xi,wi,"k.-")legend("原始点","线性点","三次样条","三次多项式")这是一点提示，自己再搞搞吧，我比较忙，要不然就帮你弄了。

int fitsBits(int x, int n) { int y = x>>(n+~1+1); return !y^!(~y);}

(1)#include <stdio.h>void main(){int i,j,n,K;char s[100],x;scanf("%d",&n);for (j=0;j<n;j++){scanf("%d ",&K);gets(s);for (i=0;i<strlen(s);i++) {x = s[i];if ( (s[i] >= "a") && (s[i] <= "z") ) {x = s[i]+K; if (x > "z") x = x - 26;if (x < "a") x = x + 26;};if ( (s[i] >= "A") && (s[i] <= "Z") ){x = s[i]+K; if (x > "Z") x = x - 26;if (x < "A") x = x + 26;}printf("%c",x); };printf(" ");};exit(0);}(2)题意不清，是不是输入数在0-15之间。也就是4bit数。否则什么叫“最高位”？只有4bit，也就是不必去做位运算，查表code[16][5]就可以了。#include <stdio.h>void main(){int d;char code[16][5]={"0000","0001","0011","0010","0110","0111","0101","0100","1100","1101","1111","1110","1010","1011","1001","1000"};printf("input n ");scanf("%d",&d);printf("%s",code[d]);}