总共有2个.cpp源文件和一个.h的头文件
related_attack.h:
void acc_x3x4(int x1[],int x2[],int x3[],int x4[],int x[]);

1.cpp:
/*  fx=x1+x2时符合率最大 P=0.75
    LFSR1: x7  x5  x3  x  1
    LFSR2: x8  x7  x2  x  1
    LFSR3: x14  x12  x2  x  1
    LFSR4: x20  x3  1
    Cx=x15+x14+x13+x12+x11+x10+x8+x7+x6+x4+1
*/
#include<stdio.h>
#include<math.h>
#include "related_attack.h"
#define N 250
#define r 15
int acc_Ca(int x,int z[]);
void check(int x[],int y[],int n);
void select_m(int a[][r],int b[],int a1,int k);
void lie_f_c(int a[][r]);
void gaosi(int a[][r],int b[],int x1[],int x2[]);

int main()
{
    double p=0.75,Er,D,Tr;
    int x1[N]={0},x2[N]={0},x3[N]={0},x4[N]={0},z[N]={1,0,0,0,0,1,1,0,1,0,1,0,1,1,1,0,0,1,0,1,0,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,1,0,1,0,1,1,1,1,0,1,0,1,1,1,1,1,1,1,0,1,0,0,0,0,1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,0,0,1,0,1,0,1,0,1,1,0,0,1,1,0,0,1,1,0,0,0,1,0,0,0,0,1,1,0,1,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,1,0,1,1,0,1,0,1,1,0,0,1,0,0,0,1,1,1,0,1,1,0,0,1,1,1,1,1,1,1,0,0,1,0,0,0,0,1,0,0,0,1,1,1,0,0,0,1,0,1,1,0,1,0,0,1,1,0,1,0,0,1,0,0,1,1,1,1,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,1,1,0,0,1,1,1,0,0,0,1,1,1,1,0,1,0,0,1,1,1,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0};
    int i,ca;
    Er=N*(2*p-1);
    D=4*N*p*(1-p);
    Tr=Er-3.09*sqrt(D);
    for (i=31976;i<32768;i++)//31977
    {
        ca=abs(acc_Ca(i,z));
        if((double)ca>=Tr)
        {
            printf("%d    ",ca);
            printf("\n accept  %d     \n ",i);
            check(x1,x2,i);
            acc_x3x4(x1,x2,x3,x4,z);//当i=31977调试完这一步时，p,Er,D,Tr,的值全部变成了0
        }
    }
   
}


int acc_Ca(int x,int z[])
{
    int i,j;
    int a[N]={0},c[r]={1,0,0,0,1,0,1,1,1,0,1,1,1,1,1};        //Cx=x15+x14+x13+x12+x11+x10+x8+x7+x6+x4+1
    for (i=0;i<r;i++)
    {
        a[i]=x%2;
        x=x/2;
    }
    for (i=r;i<N;i++)
        for (j=0;j<r;j++)
            a[i]=(a[i]+a[i-r+j]*c[j])%2;
    for (j=0,i=0;i<N;i++)
    {
        j=j+1-2*abs((a[i]+z[i])%2);
    }
    return j;
}


void check (int x[],int y[],int n)
{
    int i;
    int c[r],a[r][r]={0};
    for (i=0;i<r;i++)
    {
        c[i]=n%2;
        n=n/2;
        printf("%d",c[i]);
    }
    printf("\n");
    lie_f_c(a);
    gaosi(a,c,x,y);
}


void lie_f_c(int a[][r])
{   
    int i,j;
    int a1[r][7]={0},a2[r][8]={0};
    for(i=0;i<r;i++){
        for(j=0;j<7;j++){
            if(i<7) a1[i][i]=1;
            else a1[i][j]=(abs(a1[i-7][j]+a1[i-6][j]+a1[i-4][j]+a1[i-2][j]))%2;
            //printf("%d",a1[i][j]);
        }
        //printf("\n");
    }
    for(i=0;i<r;i++){
        for(j=0;j<8;j++){
            if(i<8) a2[i][i]=1;
            else a2[i][j]=(abs(a2[i-8][j]+a2[i-7][j]+a2[i-6][j]+a2[i-1][j]))%2;
            //printf("%d",a2[i][j]);
        }
        //printf("\n");
    }
    for(i=0;i<r;i++){
        for(j=0;j<r;j++){
            if(j<7) a[i][j]=a1[i][j];
            else a[i][j]=a2[i][j-7];
            //printf("%d",a[i][j]);
        }
        //printf("\n");
    }
}

/*高斯消元法解线性方程组*/
void gaosi(int a[][r],int b[],int x1[],int x2[])
{
    int k,j,i;
    for(k=0;k<r;k++){
        select_m(a,b,a[k][k],k);//选取主元素
        for(j=k+1;j<r;j++){
            a[k][j]=(abs(a[k][j]/a[k][k]))%2;     //wrong!!
        }
        b[k]=(b[k]/a[k][k])%2;
        for(i=k+1;i<r;i++){
            for(j=k+1;j<r;j++){
                a[i][j]=(abs(a[i][j]-a[i][k]*a[k][j]))%2;
            }
            b[i]=(abs(b[i]-a[i][k]*b[k]))%2;
        }
    }
    for(i=r-2;i>=0;i--){
        for(j=i+1;j<r;j++){
            b[i]=(abs(b[i]-a[i][j]*b[j]))%2;
        }
    }
    for(i=0;i<r;i++){
        if(i<7) {
            if(i==0)
                printf("序列a的初态： ");
            x1[i]=b[i];
            printf("%d ",b[i]);
        }
        else {
            if(i==7)
                printf("\n序列b的初态： ");
            x2[i-7]=b[i];
            printf("%d ",b[i]);
        }
    }
    printf("\n");
}

/*选主元素*/
void select_m(int a[][r],int b[],int a1,int k)
{
    int d,t;
    int l,i,j;
    d=a1;
    l=k;
    for(i=k+1;i<r;i++){
        if(abs(a[i][k])>abs(d)){
            d=a[i][k];
            l=i;
        }
    }
    if(d==0)
        printf("fail!!");
    else
        if(l!=k){
            for(j=k;j<r;j++){
                t=a[l][j];
                a[l][j]=a[k][j];
                a[k][j]=t;
            }
            t=b[k];
            b[k]=b[l];
            b[l]=t;
        }
}
2.cpp:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include "related_attack.h"
#define N 250
void acc(int x,int p,int n,int x1[],int x2[],int x3[],int x4[],int z[],int y[]);
void acc_x1(int x[]);
void acc_x2(int x[]);
void acc_x3(int x[]);
void acc_x4(int x1[],int x2[],int x3[],int x4[],int z[]);
int g(int x,int y,int z,int w);
int ensure(int x1[],int x2[],int x3[],int x4[],int z[]);

void acc_x3x4(int x1[],int x2[],int x3[],int x4[],int x[])//x是z
{
    int b[14],k[14]={0};
    int i,j,n,p=0;
    acc_x1(x1);
    acc_x2(x2);
    for(i=0,j=0;i<14;i++)    //找可以由x1x2直接确定的x3x4
    {
        if(x1[i]==0&&x2[i]==0&&x[i]==1)
        {
            x3[i]=1;
            x4[i]=1;
            k[j]=i;
            j++;
        }
        else if(x1[i]==0&&x2[i]==1&&x[i]==0)
        {
            x3[i]=0;
            x4[i]=1;
            k[j]=i;
            j++;
        }
        else if(x1[i]==1&&x2[i]==0&&x[i]==0)
        {
            x3[i]=1;
            x4[i]=0;
            k[j]=i;
            j++;
        }
        else if(x1[i]==1&&x2[i]==1&&x[i]==1)
        {
            x3[i]=1;
            x4[i]=1;
            k[j]=i;
            j++;
        }
        
    }
/*    printf("%d\n",j);
    for (i=0;i<14;i++)            //调试
        printf("%4d",x1[i]);
    printf("\n");
    for (i=0;i<14;i++)
        printf("%4d",x2[i]);
    printf("\n");                //调试
    for (i=0;i<14;i++)
        printf("%4d",x[i]);
    printf("\n");
    for (i=0;i<14;i++)
        printf("%4d",k[i]);
    printf("\n");    */
    n=(int)pow(2.0,(double)(14-j));
    for(i=0;i<n;i++)
    {
        acc(i,p,n,x1,x2,x3,x4,x,k);    //p用作计数来判断x3x4是否合适
        if(p==250)
        {
            for(i=0;i<N;i++)
                if(i==0)
                    printf("x1序列：%d",x1[i]);
                else
                    printf("%d",x1[i]);
            printf("\n");
            for(i=0;i<N;i++)
                if(i==0)
                    printf("x2序列：%d",x2[i]);
                else
                    printf("%d",x2[i]);
            printf("\n");
            for(i=0;i<N;i++)
                if(i==0)
                    printf("x3序列：%d",x3[i]);
                else
                    printf("%d",x3[i]);
            printf("\n");
            for(i=0;i<N;i++)
                if(i==0)
                    printf("x4序列：%d",x4[i]);
                else
                    printf("%d",x4[i]);
            printf("\n");
        }
    }
}


    void acc (int x,int p,int n,int x1[],int x2[],int x3[],int x4[],int z[],int y[])
    {
        
        int i,j,k,m=0,q=x;
        for(i=0,j=0;i<14;i++)
        {
            if(i==y[j])
                j++;
            else
            {
                x3[i]=q%2;
                q=q/2;
            }
        }
        
        for(k=0;k<n;k++,m=k)
        {
            for(i=0,j=0;i<n;i++)//当i=260时，x3[7]由1变为9；当i=261时，x3[7]和x3[8]都由1变为0；当i=264时，x3[7]x3[8]x3[11]都由1变为0
            {
                if(i==y[j])
                    j++;
                else
                {
                    x4[i]=m%2;
                    m=m/2;
                }
            }
            acc_x3(x3);
            acc_x4(x1,x2,x3,x4,z);
            p=ensure(x1,x2,x3,x4,z);
//            printf("%4d",p);
            if(p==250)
                break;
        }
    }

    void acc_x1(int x[])
    {
        int i;
        for(i=7;i<N;i++)
        {   
            x[i]=(x[i-7]+x[i-6]+x[i-4]+x[i-2])%2;
        }
    }// x7  x5  x3  x  1

    void acc_x2(int x[])
    {
        int i;
        for(i=8;i<N;i++)
            x[i]=(x[i-8]+x[i-7]+x[i-6]+x[i-1])%2;
    }// x8  x7  x2  x  1

    void acc_x3(int x[])
    {
        int i;
        for(i=14;i<N;i++)
            x[i]=(x[i-2]+x[i-12]+x[i-13]+x[i-14])%2;

    } //x14  x12  x2  x  1

    void acc_x4(int x1[],int x2[],int x3[],int x4[],int z[])
    {
        int i,j;
        for(i=14;i<20;i++)
            if(z[i]==g(x1[i],x2[i],x3[i],0))
                    x4[i]=0;
                else
                    x4[i]=1;
        for(i=20;i<N;i++)
            x4[i]=(x4[i-17])%2;
    }// x20  x3

    int g(int x,int y,int z,int w)
    {
        int i;
        i=(x+y+x*z+y*w+z*w+x*y*z+x*y*w)%2;
        return i;
    }//x1x2x1x3 x2x4x3x4x1x2x3x1x2x4

    int ensure(int x1[],int x2[],int x3[],int x4[],int z[])
    {
        int i;
        for(i=0;i<N;i++)
            if(z[i]!=g(x1[i],x2[i],x3[i],x4[i]))
                break;
        return i;
    }


问题就在代码的注释中，1.cpp里如果不把“acc_x3x4(x1,x2,x3,x4,z);”这一行注释掉，当i=31977调试完这一步时，p,Er,D,Tr,的值全部变成了0，如果注释掉p,Er,D,Tr,的值就一直正常；
2.cpp里 acc（int x,int p,int n,int x1[],int x2[],int x3[],int x4[],int z[],int y[]）函数里第二个循环当i=260时，x3[7]由1变为9；当i=261时，x3[7]和x3[8]都由1变为0；当i=264时，x3[7]x3[8]x3[11]都由1变为0，而且我调试以后在这个双重循环里到后面对x4[]的赋值全改变到了对x3[]赋值，不知道为什么。。

呃呃呃呃呃，来人帮忙看看吧.............
我已经无能为力了.........

坐等高手。留个记号、

楼主已经是高手了

毛线。。。写的代码基本就没完整过，从来都跑不出来，不是这里错就是那里错，还经常找不到哪里错了。。。

有人没。。。帮帮忙啊。。。

来人吧

呃。。。有点抽象，我要学多久才能看懂这个？？？

研究生楼主，我不知道你要干嘛啊。。。。。这个与你要解决的问题有哪些关系？这让大家怎么着手帮你？

lz正正宗宗破本科一个。。。。程序是干什么的一句话也说不清楚，现在只想知道我上面说的那几个地方到底是怎么回事