对整型大数的幂次求模

#include<stdio.h>
#include<string.h>
int division(int x[],int n)/*将十进制转换为二进制*/
{
int i,m,y[100],j,s,t[100];
j=0;
while(n!=0)
{
for(m=0;m<n;m++)
{
   y[m]=x[m]/2;
   x[m+1]+=(x[m]-x[m]/2*2)*10;
}
    t[j++]=x[m-1]-x[m-1]/2*2;
for(m=0;m<n;m++)x[m]=0;
    for(i=0;i<n;i++)if(y[i]!=0)break;
    n=n-i;
    for(s=0;s<n;s++)x[s]=y[i+s];
}
for(m=0;m<j/2;m++)
{
i=t[m];t[m]=t[j-1];t[j-m-1]=i;
}
for(m=0;m<j;m++)x[m]=t[m];
return j;
}
int Multiplication(int *L1,int *L2,int m,int n)
{
int i,j,t,L3[100];
for(i=0;i<100;i++)L3[i]=0;

for(i=0;i<m/2;i++)
{
t=L1[i];
    L1[i]=L1[m-i-1];
L1[m-i-1]=t;
}
for(i=0;i<n/2;i++)
{
t=L2[i];
    L2[i]=L2[n-i-1];
L2[n-i-1]=t;
}
for(j=0;j<m;j++)
{    
        t=j;
for(i=0;i<n;i++)
{
L3[t]=L3[t]+L1[j]*L2[i];
    t++;
}
}
i=m+n-1;
for(j=0;j<i;j++)
{
L3[j+1]+=L3[j]/10;
L3[j]=L3[j]%10;
}
if(L3[i]!=0)
m=i+1;
else m=i;