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=0;i<32768;i++)//31977
    {
        ca=abs(acc_Ca(i,z));
        if((double)ca>=Tr)        //只有当i=31977时才满足条件
        {
            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
        }
    }
   
}

和注释中的一样，但是如果不调试acc_x3x4(x1,x2,x3,x4,z)这一步，p,Er,D,Tr,的值还是原来的状态

#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++;
        }
        
    }
    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;
        for(i=0,j=0;i<14;i++)
        {
            if(i==y[j])
                j++;
            else
            {
                x3[i]=x%2;
                x=x/2;
            }
        }
        
        for(k=0;k<n;k++,m=k)
        {
            for(i=0,j=0;i<n;i++)
            {
                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;
    }

acc_x3x4程序有点小问题我还没找出来，但是就算是这样和p,Er,D,Tr这些值也没有关系啊，不明白为什么会变

木有关系，谢谢。不过，能不能叫几个过来帮忙一起看一下

只单纯看代码的确很头晕。。。

答题思路是这样，acc_x3x4之前都没问题，acc_x3x4求的是根据x1和x2两个250比特的数列求出x3和x4两个250比特的数列，acc_x1/2/3/4分别是根据前x比特生产后面的数据。acc主要功能是先根据x1x2确定前14比特中x3和x4中能够确定的数据，然后在非确定的地方开始穷举。但是就是这里好像有点问题