Codeforces 17D notepad simple number theory

According to the meaning of the question, ANW = (b-1) * B ^ (n-1) % C, the focus is to evaluate B ^ (n-1 ).

Weak scum cannot be pushed out and can only be publicized.

Phi (c) is less than the number of C and the number of mutual quality with C.

When x> = PHI (c): A ^ x = a (x % PHI (c) + PHI (c )).

When x <PHI (c) is used, you can directly find it.

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>#include <cmath>#include <stack>#include <map>#include <ctime>#include <iomanip>#pragma comment(linker, "/STACK:1024000000");#define EPS (1e-6)#define _LL long long#define ULL unsigned long long#define LL __int64#define INF 0x3f3f3f3f#define Mod 1000000007/** I/O Accelerator Interface .. **/#define g (c=getchar())#define d isdigit(g)#define p x=x*10+c-'0'#define n x=x*10+'0'-c#define pp l/=10,p#define nn l/=10,ntemplate<class T> inline T& RD(T &x){    char c;    while(!d);    x=c-'0';    while(d)p;    return x;}template<class T> inline T& RDD(T &x){    char c;    while(g,c!='-'&&!isdigit(c));    if (c=='-')    {        x='0'-g;        while(d)n;    }    else    {        x=c-'0';        while(d)p;    }    return x;}inline double& RF(double &x)      //scanf("%lf", &x);{    char c;    while(g,c!='-'&&c!='.'&&!isdigit(c));    if(c=='-')if(g=='.')        {            x=0;            double l=1;            while(d)nn;            x*=l;        }        else        {            x='0'-c;            while(d)n;            if(c=='.')            {                double l=1;                while(d)nn;                x*=l;            }        }    else if(c=='.')    {        x=0;        double l=1;        while(d)pp;        x*=l;    }    else    {        x=c-'0';        while(d)p;        if(c=='.')        {            double l=1;            while(d)pp;            x*=l;        }    }    return x;}#undef nn#undef pp#undef n#undef p#undef d#undef gusing namespace std;char s[1000010];int B[1000010],N[1000010];LL c,b,n;bool vis[1000010];LL prime[1000010];int Top;LL Euler(LL n){    LL ret=1,i;    for(i=2; i*i<=n; i++)    {        if(n%i==0)        {            n/=i,ret*=i-1;            while(n%i==0)                n/=i,ret*=i;        }    }    if(n>1)        ret*=n-1;    return ret;}LL Cal(int *n,LL c,bool &mark){    LL re = 0;    for(int i = 0; n[i] != -1; ++i)    {        re *= 10,re += n[i];        if(re >= c)            mark = true,re %= c;    }    return re;}LL qm(LL a,LL b,LL n){    LL ret=1;    LL tmp=a;    while(b)    {        if(b&1) ret=ret*tmp%n;        tmp=tmp*tmp%n;        b>>=1;    }    return ret;}int main(){    int i,j;    scanf("%s",s);    for(i = 0; s[i] != '\0'; ++i)        B[i] = s[i]-'0';    B[i] = -1;    scanf("%s",s);    for(i = 0; s[i] != '\0'; ++i)        N[i] = s[i]-'0';    N[i] = -1;    scanf("%I64d",&c);    LL MAXN = sqrt(c);    memset(vis,false,sizeof(vis));    Top = 0;    for(i = 2; i <= MAXN; ++i)        if(vis[i] == false)            for(j = i+i,prime[Top++] = i; j <= MAXN; j += i)                vis[j] = true;    LL phi = Euler(c),b1;    bool mark = false;    b = Cal(B,c,mark);    mark = false;    n = Cal(N,phi,mark);    b1 = (b-1+c)%c;    if(mark == true)        n = (n-1+phi)%phi + phi;    else        n--;    LL anw = b1*qm(b,n,c)%c;    if(anw == 0)        anw = c;    printf("%I64d\n",anw);    return 0;}

Codeforces 17D notepad simple number theory