Raising modulo numbers (zoj 2150)

This is the answer to the question.

#include<iostream>#include<cmath>using namespace std;int a[45002];int b[45002];int mod(int a, int b, int c){    int z = 1;    while(b)    {        if(b%2)            z = (z*a)%c;        b/=2;        a = (a*a)%c;    }    return z;}int main(){    int T;    cin>>T;    while(T--)    {        int M;        int H;        cin>>M>>H;        for(int i=0; i<H; i++)        {            cin>>a[i]>>b[i];        }        int t =0;        int mode = 0;        while(t<H)        {            int tem0 = a[t]%M;            mode = (mod(tem0,b[t],M) + mode)%M;            t++;        }        cout<<mode<<endl;    }}

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Analyze only the function for power-modulo. Let's take an example (a A a). This is B a (B = 10 ), use Z to record the generated by two-point division. Use a to replace a * A after each Binary Division. This results in a scale-down. Finally, it must be B = 1, so now we only need to use the combination of a and Z to get the final result.

int mod(int a, int b, int c){    int z = 1;    while(b)    {        if(b%2)            z = (z*a)%c;        b/=2;        a = (a*a)%c;    }    return z;}

This function is just written in a binary method, but I wrote it in the beginning.

long long F(long long a, long long b, long long m){    if(b==1)        return a%m;    else if(b%2==0)        return (F(a,b/2,m)*F(a,b/2,m))%m;    else        return (F(a,b/2,m)*F(a,b/2,m)*a%m);}

The result of this write is re. It is estimated that the recursion is too deep, leading to stack overflow. This write is very poor and does not have a binary effect at all. Instead, it takes more time than O (n.

Raising modulo numbers (zoj 2150)