算法竞赛入门10.1数论初步例题代码

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10.1 Colossal Fibonacci Numbers! UVA11582

思路：循环节+快速幂

#include<cstdio>#include<cstring>#include<algorithm>#define  LLu long long unsignedusing namespace std;inline int qpow(LLu x,LLu y ,int MOD){    x%=MOD;    LLu ans = 1,tem = x ;    while(y)    {        if(y&1)ans = (ans*tem)%MOD;        tem = (tem * tem) %MOD;        y/=2;    }return (int)ans ;}LLu A[1000000+5];//1 1 2 3 5 8 13 21inline void print(int n){    for(int i=0;i<n;i++)        printf("%llu %llu\n",A[i],A[i+1]);}int main(){    LLu a,b;    int T;scanf("%d",&T);    while(T--)    {        int n;        scanf("%llu%llu%d",&a,&b,&n);        A[0]=0%n;A[1]=1%n;int m=n*n;        for(int i=2;i<=m;i++)        {            A[i]=(A[i-1]+A[i-2])%n;            if(A[i]==A[1]&&A[i-1]==A[0])            {                n=i-1 ;                break;            }        }//print(n);        int ans = qpow(a,b,n);//printf("ans = %d\n",ans);        printf("%llu\n",A[ans]);    }}

10.3 Choose and Divide UVA10375

思路：唯一分解出所有素数个数

#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>#define __int64 long longusing namespace std;const int N = 10000+5;inline __int64 C(__int64 n,__int64 m){}int prime[2000];int num[2000l];bool judge_prime[N]={0};inline int init(){    int num = 0 ;    memset(judge_prime,false,sizeof(judge_prime));    for(int i=2;i<N;i++)    {        if(!judge_prime[i]){            prime[num++]=i;            for(int j=i*i;j<N;j+=i)                judge_prime[j]=true;        }    }return num ;}inline void add_primefactor(int n,int d,int primenum){    for(int i=0;i<primenum&&n>1;i++)    {        while(n%prime[i]==0){            num[i]+=d;            n/=prime[i];        }    }}inline void add_fun(int n,int d,int primenum){    for(int i=2;i<=n;i++)        add_primefactor(i,d,primenum);}void print(int n){    for(int i=0;i<n;i++)        printf("%d -- >  %d\n",prime[i],num[i]),getchar();}inline void work(int a,int b,int c,int d,int primenum){    add_fun(a,1,primenum);    add_fun(c,-1,primenum);    add_fun(b,-1,primenum);    add_fun(d,1,primenum);    add_fun(a-b,-1,primenum);    add_fun(c-d,1,primenum);}int main(){    int primenum = init();    int a,b,c,d;    while(scanf("%d%d%d%d",&a,&b,&c,&d)==4)    {        memset(num,0,sizeof(num));        work(a,b,c,d,primenum);        double ans = 1;        for(int i=0;i<primenum;i++)        {            ans *= pow(prime[i],num[i]);            if(prime[i]>=max(a,c))break;        }        printf("%.5f\n",ans);    }}

10.4 Minimun Sum LCM UVA10791

思路：唯一分解

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;inline int work(int a,int& x){    int res = 1;    while(x%a==0){        res*=a;        x/=a;    }return res;}int main(){    int n,t=0;    while(scanf("%d",&n)==1&&n)    {        if(n==1){            printf("Case %d: %d\n",++t,2);            continue;        }        int k=0;        long long ans = (long long)n+1;        int cur = 0;int m=n/2;        for(int i=2;i<=m&&i<=n;i++)        {            if(n%i==0){                k++;                cur += work(i,n);            }        }//printf("cur=%d\n",cur);        if(cur){            ans = min( ans , k<2?cur+1LL:cur+0LL );        }        printf("Case %d: %lld\n",++t,ans);    }}

10.5 GCD XOR UVA12716

思路：异或的基本性质：a^b=c,a^c=b + 筛素数

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int N = 30000000+5;int A[N];inline void init(){    int x = N ;    memset(A,0,sizeof(A));    for(int i=1;i<=x;i++)    {        A[i]+=A[i-1];        for(int j=i*2;j<=x;j+=i)        {            int l= j^i;            if(j>l&&(j-l)==i)                A[j]++;//printf("a=%d,b=%d,c=%d\n",j,j^i,i);        }    }}int main(){    int T,t=0;init();    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        printf("Case %d: %d\n",++t,A[n]);    }}

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