printf("%d",a[a[99]-i-1]);
printf("\n\n");
return;
}
intcmp(inta1[MAX],inta2[MAX])
{intl1,l2;
inti;
l1=a1[99];
l2=a2[99];
if(l1>l2)
return1;
if(l1return-1;
for(i=(l1-1);i>=0;i--)
{
if(a1[i]>a2[i])
return1;
if(a1[i]return-1;
}
return0;
}
voidmov(inta[MAX],int*b)
{
intj;
for(j=0;jb[j]=a[j];
return;
}
voidmul(inta1[MAX],inta2[MAX],int*c)
{
inti,j;
inty;
intx;
intz;
intw;
intl1,l2;
l1=a1[MAX-1];
l2=a2[MAX-1];
if(a1[MAX-2]=='-'&&a2[MAX-2]=='-')
c[MAX-2]=0;
elseif(a1[MAX-2]=='-')
c[MAX-2]='-';
elseif(a2[MAX-2]=='-')
c[MAX-2]='-';
for(i=0;i{
for(j=0;j{
x=a1[i]*a2[j];
y=x/10;
z=x%10;
w=i+j;
c[w]=c[w]+z;
c[w+1]=c[w+1]+y+c[w]/10;
c[w]=c[w]%10;
}
}
w=l1+l2;
if(c[w-1]==0)w=w-1;
c[MAX-1]=w;
return;
}
voidadd(inta1[MAX],inta2[MAX],int*c)
{
inti,l1,l2;
intlen,temp[MAX];
intk=0;
l1=a1[MAX-1];
l2=a2[MAX-1];
if((a1[MAX-2]=='-')&&(a2[MAX-2]=='-'))
{
c[MAX-2]='-';
}
elseif(a1[MAX-2]=='-')
{
mov(a1,temp);
temp[MAX-2]=0;
sub(a2,temp,c);
return;
}
elseif(a2[MAX-2]=='-')
{
mov(a2,temp);
temp[98]=0;
sub(a1,temp,c);
return;
}
if(l1elselen=l2;
for(i=0;i{
c[i]=(a1[i]+a2[i]+k)%10;
k=(a1[i]+a2[i]+k)/10;
}
if(l1>len)
{
for(i=len;i{
c[i]=(a1[i]+k)%10;
k=(a1[i]+k)/10;
}
if(k!
=0)
{
c[l1]=k;
len=l1+1;
}
elselen=l1;
}
else
{
for(i=len;i{
c[i]=(a2[i]+k)%10;
k=(a2[i]+k)/10;
}
if(k!
=0)
{
c[l2]=k;
len=l2+1;
}
elselen=l2;
}
c[99]=len;
return;
}
voidsub(inta1[MAX],inta2[MAX],int*c)
{
inti,l1,l2;
intlen,t1[MAX],t2[MAX];
intk=0;
l1=a1[MAX-1];
l2=a2[MAX-1];
if((a1[MAX-2]=='-')&&(a2[MAX-2]=='-'))
{
mov(a1,t1);
mov(a2,t2);
t1[MAX-2]=0;
t2[MAX-2]=0;
sub(t2,t1,c);
return;
}
elseif(a2[MAX-2]=='-')
{
mov(a2,t2);
t2[MAX-2]=0;
add(a1,t2,c);
return;
}
elseif(a1[MAX-2]=='-')
{
mov(a2,t2);
t2[MAX-2]='-';
add(a1,t2,c);
return;
}
if(cmp(a1,a2)==1)
{
len=l2;
for(i=0;i{
if((a1[i]-k-a2[i])<0)
{
c[i]=(a1[i]-a2[i]-k+10)%10;
k=1;
}
else
{
c[i]=(a1[i]-a2[i]-k)%10;
k=0;
}
}
for(i=len;i{
if((a1[i]-k)<0)
{
c[i]=(a1[i]-k+10)%10;
k=1;
}
else
{
c[i]=(a1[i]-k)%10;
k=0;
}
}
if(c[l1-1]==0)/*使得数组C中的前面所以0字符不显示了,如1000-20=0980--->显示为980了*/
{
len=l1-1;
i=2;
while(c[l1-i]==0)/*111456-111450=00006,消除0后变成了6;*/
{
len=l1-i;
i++;
}
}
else
{
len=l1;
}
}
else
if(cmp(a1,a2)==(-1))
{
c[MAX-2]='-';
len=l1;
for(i=0;i{
if((a2[i]-k-a1[i])<0)
{
c[i]=(a2[i]-a1[i]-k+10)%10;
k=1;
}
else
{
c[i]=(a2[i]-a1[i]-k)%10;
k=0;
}
}
for(i=len;i{
if((a2[i]-k)<0)
{
c[i]=(a2[i]-k+10)%10;
k=1;
}
else
{
c[i]=(a2[i]-k)%10;
k=0;
}
}
if(c[l2-1]==0)
{
len=l2-1;
i=2;
while(c[l1-i]==0)
{
len=l1-i;
i++;
}
}
elselen=l2;
}
elseif(cmp(a1,a2)==0)
{
len=1;
c[len-1]=0;
}
c[MAX-1]=len;
return;
}
voidmod(inta[MAX],intb[MAX],int*c)/*/c=amodb//注意:
经检验知道此处A和C的数组都改变了。
*/
{intd[MAX];
mov(a,d);
while(cmp(d,b)!
=(-1))/*/c=a-b-b-b-b-b.......until(c
{
sub(d,b,c);
mov(c,d);/*/c复制给a*/
}
return;
}
voiddivt(intt[MAX],intb[MAX],int*c,int*w)/*//试商法//调用以后w为amodb,C为adivb;*/
{
inta1,b1,i,j,m;/*w用于暂时保存数据*/
intd[MAX],e[MAX],f[MAX],g[MAX],a[MAX];
mov(t,a);
for(i=0;ie[i]=0;
for(i=0;id[i]=0;
for(i=0;ia1=a[MAX-1];
b1=b[MAX-1];
if(cmp(a,b)==(-1))
{
c[0]=0;
c[MAX-1]=1;
mov(t,w);
return;
}
elseif(cmp(a,b)==0)
{
c[0]=1;
c[MAX-1]=1;
w[0]=0;
w[MAX-1]=1;
return;
}
m=(a1-b1);
for(i=m;i>=0;i--)/*341245/3=341245-300000*1--->41245-30000*1--->11245-3000*3--->2245-300*7--->145-30*4=25--->25-3*8=1*/
{
for(j=0;jd[j]=0;
d[i]=1;
d[MAX-1]=i+1;
mov(b,g);
mul(g,d,e);
while(cmp(a,e)!
=(-1))
{
c[i]++;
sub(a,e,f);
mov(f,a);/*f复制给g*/
}
for(j=i;je[j]=0;
}
mov(a,w);
if(c[m]==0)c[MAX-1]=m;
elsec[MAX-1]=m+1;
return;
}
voidmulmod(inta[MAX],intb[MAX],intn[MAX],int*m)/*解决了m=a*bmodn;*/
{
intc[MAX],d[MAX];
inti;
for(i=0;id[i]=c[i]=0;
mul(a,b,c);
divt(c,n,d,m);
for(i=0;iprintf("%d",m[m[MAX-1]-i-1]);
printf("\nmlengthis:
%d\n",m[MAX-1]);
}
/*接下来的重点任务是要着手解决m=a^pmodn的函数问题。
*/
voidexpmod(inta[MAX],intp[MAX],intn[MAX],int*m)
{
intt[MAX],l[MAX],temp[MAX];/*/t放入2,l放入1;*/
intw[MAX],s[MAX],c[MAX],b[MAX],i;
for(i=0;ib[i]=l[i]=t[i]=w[i]=0;
t[0]=2;t[MAX-1]=1;
l[0]=1;l[MAX-1]=1;
mov(l,temp);
mov(a,m);
mov(p,b);
while(cmp(b,l)!
=0)
{
for(i=0;iw[i]=c[i]=0;
divt(b,t,w,c);/*//c=pmod2w=p/2*/
mov(w,b);/*//p=p/2*/
if(cmp(c,l)==0)/*/余数c==1*/
{
for(i=0;iw[i]=0;
mul(temp,m,w);
mov(w,temp);
for(i=0;iw[i]=c[i]=0;
divt(temp,n,w,c);/*/c为余c=temp%n,w为商w=temp/n*/
mov(c,temp);
}
for(i=0;is[i]=0;
mul(m,m,s);//s=a*a
for(i=0;ic[i]=0;
divt(s,n,w,c);/*/w=s/n;c=smodn*/
mov(c,m);
}
for(i=0;is[i]=0;
mul(m,temp,s);
for(i=0;ic[i]=0;
divt(s,n,w,c);
mov(c,m);/*余数s给m*/
m[MAX-2]=a[MAX-2];/*为后面的汉字显示需要,用第99位做为标记*/
return;
/*/k=temp*k%n;*/
}
intis_prime_san(intp[MAX])
{
inti,a[MAX],t[MAX],s[MAX],o[MAX];
for(i=0;is[i]=o[i]=a[i]=t[i]=0;
t[0]=1;
t[MAX-1]=1;
a[0]=2;//{2,3,5,7}
a[MAX-1]=1;
sub(p,t,s);
expmod(a,s,p,o);
if(cmp(o,t)!
=0)
{
return0;
}
a[0]=3;
for(i=0;iexpmod(a,s,p,o);
if(cmp(o,t)!
=0)
{
return0;
}
a[0]=5;
for(i=0;iexpmod(a,s,p,o);
if(cmp(o,t)!
=0)
{
return0;
}
a[0]=7;
for(i=0;iexpmod(a,s,p,o);
if(cmp(o,t)!
=0)
{
return0;
}
return1;
}
intcoprime(inte[MAX],ints[MAX])/*////求两个大数之间是否互质////*/
{
inta[MAX],b[MAX],c[MAX],d[MAX],o[MAX],l[MAX];
inti;
for(i=0;il[i]=o[i]=c[i]=d[i]=0;
o[0]=0;o[MAX-1]=1;
l[0]=1;l[MAX-1]=1;
mov(e,b);
mov(s,a);
do
{
if(cmp(b,l)==0)
{
return1;
}
for(i=0;ic[i]=0;
divt(a,b,d,c);
mov(b,a);/*b--->a*/
mov(c,b);/*c--->b*/
}while(cmp(c,o)!
=0);
/*printf("Iheyarenotcoprime!
\n");*/
return0;
}
voidprime_random(int*p,int*q)
{
inti,k;
time_tt;
p[0]=1;
q[0]=3;
//p[19]=1;
//q[18]=2;
p[MAX-1]=10;
q[MAX-1]=11;
do
{
t=time(NULL);
srand((unsignedlong)t);
for(i=1;i
{
k=rand()%10;
p[i]=k;
}
k=rand()%10;
while(k==0)
{
k=rand()%10;
}
p[p[MAX-1]-1]=k;
}while((is_prime_san(p))!
=1);
printf("素数p为:
");
for(i=0;i
{
printf("%d",p[p[MAX-1]-i-1]);
}
printf("\n\n");
do
{
t=time(NULL);
srand((unsignedlong)t);
for(i=1;i{
k=rand()%10;
q[i]=k;
}
}while((is_prime_san(q))!
=1);
printf("素数q为:
");
for(i=0;i{
printf("%d",q[q[MAX-1]-i-1]);
}
printf("\n\n");
return;
}
voiderand(inte[MAX],intm[MAX])
{
inti,k;
time_tt;
e[MAX-1]=5;
printf("随机产生一个与(p-1)*(q-1)互素的e:
");
do
{
t=time(NULL);
srand((unsignedlong)t);
for(i=0;i{
k=rand()%10;
e[i]=k;
}
while((k=rand()%10)==0)
k=rand()%10;
e[e[MAX-1]-1]=k;
}while(coprime(e,m)!
=1);
for(i=0;i{
printf("%d",e[e[MAX-1]-i-1]);
}
printf("\n\n");
return;
}
voidrsad(inte[MAX],intg[MAX],int*d)
{
intr[MAX],n1[MAX],n2[MAX],k[MAX],w[MAX];
inti,t[MAX],b1[MAX],b2[MAX],temp[MAX];
mov(g,n1);
mov(e,n2);
for(i=0;ik[i]=w[i]=r[i]=temp[i]=b1[i]=b2[i]=t[i]=0;
b1[MAX-1]=0;b1[0]=0;/*/b1=0;*/
b2[MAX-1]=1;b2[0]=1;/*/b2=1;*/
while
(1)
{
for(i=0;ik[i]=w[i]=0;
divt(n1,n2,k,w);/*/k=n1/n2;*/
for(i=0;itemp[i]=0;
mul(k,n2,temp);/*/temp=k*n2;*/
for(i=0;ir[i]=0;
sub(n1,temp,r);
if((r[MAX-1]==1)&&(r[0]==0))/*/r=0*/
{
break;
}
else
{
mov(n2,n1);/*/n1=n2;*/
mov(r,n2);/*/n2=r;*/
mov(b2,t);/*/t=b2;*/
for(i=0;itemp[i]=0;
mul(k,b2,temp);/*/b2=b1-k*b2;*/
for(i=0;ib2[i]=0;
sub(b1,temp,b2);
mov(t,b1);
}
}
for(i=0;it[i]=0;
add(b2,g,t);
for(i=0;itemp[i]=d[i]=0;
divt(t,g,temp,d);
printf("由以上的(p-1)*(q-1)和e计算得出的d:
");
for(i=0;iprintf("%d",d[d[MAX-1]-i-1]);
printf("\n\n");
}
/*/求解密密钥d的函数(根据Euclid算法)96403770511368768000*/
unsignedlongrsa(unsignedlongp,unsignedlongq,unsignedlonge)/*/求解密密钥d的函数(根据Euclid算法)*/
{
unsignedlongg,k,r,n1,n2,t;
unsignedlongb1=0,b2=1;
g=(p-1)*(q-1);
n1=g;
n2=e;
while
(1)
{
k=n1/n2;
r=n1-k*n2;
if(r!
=0)
{
n1=n2;
n2=r;
t=b2;
b2=b1-k*b2;
b1=t;
}
else
{
break;
}
}
return(g+b2)%g;
}
/*/------------------------------------------导入导出公钥和私钥------------------------------------/*/
voidloadpkey(inte[MAX],intn[MAX])//导入公钥
{
FILE*fp;
charfilename[25],str[MAX],ch;
inti,k;
for(i=0;ie[i]=n[i]=0;
while
(1)
{
printf("\n"