A5算法C语言实现报告（推荐）_a5算法c语言实现报告

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程序：

#include #define N 64

lfsr(int a,int b,int c,int d,int T[]);void main(){ int A[19]={1,0,1,1,0,0,1,1,0,0,0,0,1,0,1,0,1,0,1};

int B[22]={0,0,1,0,0,0,1,1,1,0,0,1,0,0,1,0,1,0,1,0,1,1};int C[23]={1,0,0,0,1,0,1,1,0,1,0,0,1,0,0,0,1,0,1,0,1,0,1};for(int i=0;i

{

printf(“%d”,A[18]^B[21]^C[22]);

lfsr(13,16,17,18,A);

lfsr(12,16,20,21,B);

lfsr(17,18,21,22,C);} else if(j==1)

{

printf(“%d”,A[18]^B[21]^C[22]);

if(A[9]==0)

lfsr(13,16,17,18,A);

if(B[11]==0)

lfsr(12,16,20,21,B);

if(C[11]==0)

lfsr(17,18,21,22,C);} else if(j==2)

{

printf(“%d”,A[18]^B[21]^C[22]);

if(A[9]==1)

lfsr(13,16,17,18,A);

if(B[11]==1)

lfsr(12,16,20,21,B);

if(C[11]==1)

lfsr(17,18,21,22,C);} else if(j==3)

{

printf(“%d”,A[18]^B[21]^C[22]);

lfsr(13,16,17,18,A);

lfsr(12,16,20,21,B);

lfsr(17,18,21,22,C);

} } printf(“n n”);} lfsr(int a,int b,int c,int d,int T[]){

int i;

for(i=d;i>0;i--)

{

T[i]=T[i-1];

}

T[0]=T[a]^T[b]^T[c]^T[d];

return(T[0]);} 密钥流：

A[19]={1,0,1,1,0,0,1,1,0,0,0,0,1,0,1,0,1,0,1};

B[22]={0,0,1,0,0,0,1,1,1,0,0,1,0,0,1,0,1,0,1,0,1,1};C[23]={1,0,0,0,1,0,1,1,0,1,0,0,1,0,0,0,1,0,1,0,1,0,1};密钥流序列：

1111 1111 1101 1010 1101 0101 0111 0001 1110 0000 1010 1000 0101 1100 0100 0000