uva 10956

首次學習了這麼快速的判斷質數方法

這個就是 費瑪測試 只要判斷這三個底數{2,7,61}都為true

就能正確判斷2^32以下的數字是質數了

複雜度為O(logN)

//
//        GGGGGGGGGGGGG        CCCCCCCCCCCCC               AAA
//     GGG::::::::::::G     CCC::::::::::::C              A:::A
//   GG:::::::::::::::G   CC:::::::::::::::C             A:::::A
//  G:::::GGGGGGGG::::G  C:::::CCCCCCCC::::C            A:::::::A
// G:::::G       GGGGGG C:::::C       CCCCCC           A:::::::::A
//G:::::G              C:::::C                        A:::::A:::::A
//G:::::G              C:::::C                       A:::::A A:::::A
//G:::::G    GGGGGGGGGGC:::::C                      A:::::A   A:::::A
//G:::::G    G::::::::GC:::::C                     A:::::A     A:::::A
//G:::::G    GGGGG::::GC:::::C                    A:::::AAAAAAAAA:::::A
//G:::::G        G::::GC:::::C                   A:::::::::::::::::::::A
// G:::::G       G::::G C:::::C       CCCCCC    A:::::AAAAAAAAAAAAA:::::A
//  G:::::GGGGGGGG::::G  C:::::CCCCCCCC::::C   A:::::A             A:::::A
//   GG:::::::::::::::G   CC:::::::::::::::C  A:::::A               A:::::A
//     GGG::::::GGG:::G     CCC::::::::::::C A:::::A                 A:::::A
//        GGGGGG   GGGG        CCCCCCCCCCCCCAAAAAAA                   AAAAAAA
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
#include <ctime>
using namespace std;
#ifdef DEBUG
#define VAR(a,b) decltype(b) a=(b)
#define debug(...) printf("DEBUG: "),printf(__VA_ARGS__)
#define gettime() end_time=clock();printf("now running time is %.7f\n",(float)(end_time - start_time)/CLOCKS_PER_SEC);
#else
#define VAR(a,b) __typeof(b) a=(b)
#define debug(...)
#define gettime()
#endif
typedef unsigned int uint;
typedef unsigned long long int Int;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\n")
#define I_de(x,n)for(int i=0;i<n;i++)printf("%d ",x[i]);Pln()
#define De(x)printf(#x"%d\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\
    printf("%d ",dp[htx][hty]);}Pln();}
#define M 55
#define PII pair<int,int>
#define PB push_back
#define oo INT_MAX
#define Set_oo 0x3f
#define FOR(a,b) for(VAR(a,(b).begin());a!=(b).end();++a)
#define eps 1e-6
clock_t start_time=clock(), end_time;
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
    int tmp=min(x,y);
    return min(tmp,z);
}
int max3(int x,int y,int z){
    int tmp=max(x,y);
    return max(tmp,z);
}
Int powmod(Int b,Int u,Int n){
    Int ans=1;
    Int base=b;
    while(u){
        if(u&1)
            ans=ans*base%n;
        base=base*base%n;
        u>>=1;
    }
    return ans;
}
bool suspect(Int b,Int n){
    Int exp=0;
    Int t=n-1;
    while(!(t&1)){
        exp++;
        t>>=1;
    }
    Int xi=powmod(b,t,n);
    if(xi==1||xi==n-1)return true;
    for(int i=0;i<exp-1;i++){
        xi=xi*xi%n;
        if(xi==1)return false;
        if(xi==n-1)return true;
    }
    return false;

}
Int b,mx,mn;
int main() {
    ios_base::sync_with_stdio(0);
//    putchar(suspect(3,121)?'y':'n');
    int ff=0;
    while(~scanf("%llu%llu%llu",&b,&mn,&mx)&&b+mn+mx){
        Int omn=mn,omx=mx;
        vector<Int> ans;
        int op=0;
        if(!(mn&1))mn++;
        if(!(mx&1))mx--;
        for(Int i=mn;i<=mx;i+=2){
//            printf("%lld\n",i);
            bool isp=false;
            if(suspect(2,i)&&suspect(7,i)&&suspect(61,i)){
                isp=true;
            }else{
                op++;
            }
            if((i&1)&&suspect(b,i)==true&&isp==false){
                ans.PB(i);
            }
        }
        int size=ans.size();
        if(ff++)Pln();
        printf("There are %d odd non-prime numbers between %llu and %llu.\n",op,omn,omx);
        if(size){
            printf("%d suspects fail in base %llu:\n",size,b);
            FOR(it,ans){
                printf("%lld\n",*it);
            }
        }else printf("There are no failures in base %llu.\n",b);
    }
















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