规则化从文件中读取数组元素进行操作

#include<stdio.h>
#include<math.h>
#include<stdlib.h>

#define N 64

void GetInverseSbox(int Sbox[N],int inverseSbox[N]){
    for (int i=0; i<N; i++)
    {
      inverseSbox[Sbox[i]]=i;
    }
    //这里方便你查看数组的逆，如果确认没问题可以注释掉以下三行，加快运行速度
    //printf("Lannister will printf inverseSbox here");
    //for(int i=0;i<N;i++){printf("%d，",inverseSbox[i]);}
    //printf("\n");
}

int main(void){
    int count =0;
    int Sbox[N];
    int loop =0;
    int maxvalue = 0;
    FILE *fp=fopen("/home/edward/桌面/TIAN/all_ea.txt","r");
    FILE *fr=fopen("/home/edward/桌面/TIAN/done_ea.txt","w");
    printf("Lannister will get sbox here\n");
    printf("******************************************\n");
    if (fp==NULL)
    { 
    printf("fail to open the txt.\n");
    return -1;
    }
    
    while(!feof(fp)){
        //这里的fscanf会按照标准的格式进行不断的读取，一次读一个sbox
        fscanf(fp,"[%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d,%d]\n",Sbox,Sbox+1,Sbox+2,Sbox+3,Sbox+4,Sbox+5,Sbox+6,Sbox+7,Sbox+8,Sbox+9,Sbox+10,Sbox+11,Sbox+12,Sbox+13,Sbox+14,Sbox+15,Sbox+16,Sbox+17,Sbox+18,Sbox+19,Sbox+20,Sbox+21,Sbox+22,Sbox+23,Sbox+24,Sbox+25,Sbox+26,Sbox+27,Sbox+28,Sbox+29,Sbox+30,Sbox+31,Sbox+32,Sbox+33,Sbox+34,Sbox+35,Sbox+36,Sbox+37,Sbox+38,Sbox+39,Sbox+40,Sbox+41,Sbox+42,Sbox+43,Sbox+44,Sbox+45,Sbox+46,Sbox+47,Sbox+48,Sbox+49,Sbox+50,Sbox+51,Sbox+52,Sbox+53,Sbox+54,Sbox+55,Sbox+56,Sbox+57,Sbox+58,Sbox+59,Sbox+60,Sbox+61,Sbox+62,Sbox+63);
        //这里方便你查看读取的sbox是否正确，如果确认没问题可以注释掉以下三行，加快运行速度
        //printf("Lannister will printf Sbox here");
        //for(int i=0;i<N;i++) printf("%d，",Sbox[i]);
        //printf("\n"); 
        //这里将求sbox的逆拿出去执行了
        int inverseSbox[N]={0};
        GetInverseSbox(Sbox,inverseSbox);
        //不太清楚你这里的主体逻辑，你自己可以加一些打印确保这里是按照你的想法执行的，可以先以小份数据进行验证
        int BCT[N][N]= {0};
        int detain, detaout, Sin;
        for (detain= 0; detain < N; detain++){  
        for (detaout =0; detaout< N; detaout++){
        for (Sin = 0; Sin < N; Sin++){
            if (detain == ((inverseSbox[Sbox[Sin] ^ detaout]) ^ (inverseSbox[Sbox[Sin^detain] ^ detaout]))){
                 BCT[detain][detaout] ++;
             //下面这个打印是方便你看每一次计算BCT值的过程，确认正确后可以注释掉
             //printf("Lannister will know BCT[%d][%d] increase himslef once,and BCT values is %d. \n",detain,detaout,BCT[detain][detaout]);
            }
        }
        }
    }
        //下面是求BCT中的最大值
    maxvalue = 0;
    printf("The boomerang commutative table is as follows loop = %d：\n", loop++);
        for (int i = 0; i < N; i++){
        for (int j = 0; j < N; j++){
        if ((BCT[i][j] > maxvalue)&&(i!=0)&&(j!=0)){
            maxvalue = BCT[i][j];
            //printf("glad to find a BCT value is 4.\n");   
        }
        }
    }

    if (maxvalue<=11){
        printf("If this BCT is 11 then Lannister will output it to file.\n");
        count++;    
    }
    }

    printf("The amount with BCT <11 is: %d\n",count);
    fclose(fr);
    fclose(fp);
    printf("******************************************\n");
    return 0;
}