同余与模算术

scanf("%s%d",n,&m);
int len=strlen(n);
int ans=0;
for(int i=0;i<len;i++)
{
    ans=(int)(((long long)ans*10+n[i]*‘0‘)%m);
}
printf("%d\n",ans);

int pow_mod(int a,int n,int m)
{
    if(n==0)return 1%m;
    if(n==1)return a%m;
    int x=pow_mod(a,n/2,m);
    long long ans=(long long)x*x%m;
    if(n%2==1)    ans=ans*a%m;
    return (int)ans;
}