beforekey[i][j]=C[i][j]; for(k=0;k<28;k++,j++)
beforekey[i][j]=D[i][k]; }
/*beforekey[16][56]经过pc_2置换，生成16轮子密钥key[16][48]*/ int keys[16][48]; for(i=0;i<16;i++) for(j=0;j<48;j++)
keys[i][j]=beforekey[i][pc_2table[j]-1]; /*初始置换IP*/
int IP_table[64]={
58,50,42,34,26,18,10,2, 60,52,44,36,28,20,12,4, 62,54,46,38,30,22,14,6, 64,56,48,40,32,24,16,8, 57,49,41,33,25,17,9,1, 59,51,43,35,27,19,11,3, 61,53,45,37,29,21,13,5, 63,55,47,39,31,23,15,7}; int afterIP[64]; for(i=0;i<64;i++) {
afterIP[i]=miwenB[IP_table[i]-1]; }
/*明文的第零次分组，L0[32],R0[32]*/ int L0[32],R0[32]; for(i=0;i<32;i++) L0[i]=afterIP[i];
for(j=0,i=32;i<64;i++,j++) R0[j]=afterIP[i];
/*十六轮循环生成密文，首先是L[16][32]和R[16][32]的生成*/ int L[16][32],R[16][32]; int E_table[48]={
32,1,2,3,4,5, 4,5,6,7,8,9,
8,9,10,11,12,13, 12,13,14,15,16,17, 16,17,18,19,20,21, 20,21,22,23,24,25, 24,25,26,27,28,29, 28,29,30,31,32,31}; int afterER0[48],beforeS[48]; for(i=0;i<32;i++) //Li=R(i-1) L[0][i]=R0[i];
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for(i=0;i<48;i++) //E置换，生成48位数据，与第十五轮子密钥进行异或，为进入S盒做准备
afterER0[i]=R0[E_table[i]-1]; for(i=0;i<48;i++)
if(afterER0[i]==keys[15][i]) beforeS[i]=0; else
beforeS[i]=1;
/*将beforeS[48]分在8个数组中，为经过8个S盒准备*/ int beforeSBox[8][6]; k=0;
for(i=0;i<8;i++) for(j=0;j<6;j++) {
beforeSBox[i][j]=beforeS[k]; k++; }
/* S盒变换，48位数据生成32位数据*/ int afterSBox[8];
int SBox[8][4][16]={ //S盒变换表 14,4,13,1,2,15,11,8,3,10,6,12,5,9,0,7, 0,15,7,4,14,2,13,1,10,6,12,11,9,5,3,8, 4,1,14,8,13,6,2,11,15,12,9,7,3,10,5,0, 15,12,8,2,4,9,1,7,5,11,3,14,10,0,6,13, 15,1,8,14,6,11,3,4,9,7,2,13,12,0,5,10, 3,13,4,7,15,2,8,14,12,0,1,10,6,9,11,5, 0,14,7,11,10,4,13,1,5,8,12,6,9,3,2,15, 13,8,10,1,3,15,4,2,11,6,7,12,0,5,14,9, 10,0,9,14,6,3,15,5,1,13,12,7,11,4,2,8, 13,7,0,9,3,4,6,10,2,8,5,14,12,11,15,1, 13,6,4,9,8,15,3,0,11,1,2,12,5,10,14,7, 1,10,13,0,6,9,8,7,4,15,14,3,11,5,2,12, 7,13,14,3,0,6,9,10,1,2,8,5,11,12,4,15, 13,8,11,5,6,15,0,3,4,7,2,12,1,10,14,9, 10,6,9,0,12,11,7,13,15,1,3,14,5,2,8,4, 3,15,0,6,10,1,13,8,9,4,5,11,12,7,2,14, 2,12,4,1,7,10,11,6,8,5,3,15,13,0,14,9, 14,11,2,12,4,7,13,1,5,0,15,10,3,9,8,6, 4,2,1,11,10,13,7,8,15,9,12,5,6,3,0,14, 11,8,12,7,1,14,2,13,6,15,0,9,10,4,5,3, 12,1,10,15,9,2,6,8,0,13,3,4,14,7,5,11, 10,15,4,2,7,12,9,5,6,1,13,14,0,11,3,8, 9,14,15,5,2,8,12,3,7,0,4,10,1,13,11,6, 4,3,2,12,9,5,15,10,11,14,1,7,6,0,8,13,
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4,11,2,14,15,0,8,13,3,12,9,7,5,10,6,1, 13,0,11,7,4,9,1,10,14,3,5,12,2,15,8,6, 1,4,11,13,12,3,7,14,10,15,6,8,0,5,9,2, 6,11,13,8,1,4,10,7,9,5,0,15,14,2,3,12, 13,2,8,4,6,15,11,1,10,9,3,14,5,0,12,7, 1,15,13,8,10,3,7,4,12,5,6,11,0,14,9,2, 7,11,4,1,9,12,14,2,0,6,10,13,15,3,5,8, 2,1,14,7,4,10,8,13,15,12,9,0,3,5,6,11}; int hang,lie; for(i=0;i<8;i++) {
hang=beforeSBox[i][0]*2+beforeSBox[i][5];
lie=beforeSBox[i][1]*8+beforeSBox[i][2]*4+beforeSBox[i][3]*2+beforeSBox[i][4];
afterSBox[i]=SBox[i][hang][lie]; }
/*将从S盒中出来的8个十进制数字转换成32位的二进制数*/ int beforep[32]; k=0;
int temp; int str1[4]; for(i=0;i<8;i++) {
temp=afterSBox[i]; for(j=0;j<4;j++) {
str1[j]=temp%2; temp=temp/2; }
for(j=3;j>=0;j--) {
beforep[k]=str1[j]; k++; } }
/*从S盒输入的32位的二进制数进行P置换，生成的即是f函数的结果*/ int p_table[32]={
16,7,20,21, 29,12,28,17, 1,15,23,26, 5,18,31,10, 2,8,24,14, 32,27,3,9, 19,13,30,6,
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22,11,4,25}; int p[32];
for(i=0;i<32;i++)
p[i]=beforep[p_table[i]-1];
/*p[32]与L0[32]进行异或，得到R[0][32]*/ for(i=0;i<32;i++) {
R[0][i]=p[i]+L0[i]; if(R[0][i]==2) R[0][i]=0; }
/*已经生成L[0][32]和R[0][32],下面进行十五轮循环，生成其余L[i][32]和R[i][32]*/
int afterER[16][48]; int n;
for(i=1;i<16;i++) {
for(j=0;j<32;j++)
L[i][j]=R[i-1][j]; for(j=0;j<48;j++) //E置换
afterER[i][j]=R[i-1][E_table[j]-1]; for(j=0;j<48;j++) //与子密钥进行异或 {
if(afterER[i][j]==keys[15-i][j]) beforeS[j]=0; else
beforeS[j]=1; } k=0;
for(j=0;j<8;j++) //生成进入S盒钱的8个数组，分别进入8个S盒 for(n=0;n<6;n++) {
beforeSBox[j][n]=beforeS[k]; k++; }
for(j=0;j<8;j++) //进入S盒，从8个S盒中生成8个数 {
hang=beforeSBox[j][0]*2+beforeSBox[j][5];
lie=beforeSBox[j][1]*8+beforeSBox[j][2]*4+beforeSBox[j][3]*2+beforeSBox[j][4];
afterSBox[j]=SBox[j][hang][lie]; } k=0;
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for(j=0;j<8;j++) //S盒出来的8个十进制数转换成32位的二进制数 {
temp=afterSBox[j]; for(n=0;n<4;n++) {
str1[n]=temp%2; temp=temp/2; }
for(n=3;n>=0;n--) {
beforep[k]=str1[n]; k++; } }
for(j=0;j<32;j++)
p[j]=beforep[p_table[j]-1]; for(j=0;j<32;j++)
if(p[j]==L[i-1][j]) R[i][j]=0; else
R[i][j]=1; }
/*通过十五轮循环，最后得到L[15][32]和R[15][32],合并之后进行IP逆置换生成明文*/
int beforeoutput[64]; int output[64]; int IPR_tabel[64]={
40,8,48,16,56,24,64,32, 39,7,47,15,55,23,63,31, 38,6,46,14,54,22,62,30, 37,5,45,13,53,21,61,29, 36,4,44,12,52,20,60,28, 35,3,43,11,51,19,59,27, 34,2,42,10,50,18,58,26, 33,1,41,9,49,17,57,25}; for(i=0;i<32;i++) //数据合并,R[15][32]在前，L[15][32]在后面 beforeoutput[i]=R[15][i]; for(i=32,j=0;i<64;i++,j++) beforeoutput[i]=L[15][j]; for(i=0;i<64;i++)
output[i]=beforeoutput[IPR_tabel[i]-1];//得到64位明文 m_Mingwen=\ for(i=0;i<64;i++) {
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