kuangbin簡單數論(上)

原創 小哧 2020-07-05 14:23

kuangbin14數論簡單題

[A - Bi-shoe and Phi-shoe]

//my cpp
#include<cstdio>
#include<cmath>
using namespace std;

inline bool inprime(int x){
    if(x<=2)return 1;
    for(int i = 2; i <= sqrt(x)+1; i++){
        if(x%i == 0)return 0;
    }
    return 1;
}

int main(){
    int t=0,T;
    scanf("%d",&T);
    while(t++<T){
        int n;
        long long ans =0;
        scanf("%d",&n);
        for(int i = 0; i < n; i++){
            int temp;
            scanf("%d",&temp);
            while(temp++){
                if(inprime(temp)){ans+=temp;break;}
            }
        }
        printf("Case %d: %lld Xukha\n",t,ans);
    }
    return 0;
}
//dabiao
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 1000000;
int prime[maxn/10],sgn[maxn+7],cnt;
inline void initial(){
    for(int i = 2; i < maxn + 7; i++){
        if(!sgn[i]){
            prime[cnt++] = i;
            for(int j = i+i; j < maxn +7; j+=i){
                sgn[j] = 1;
            }
        }
    }
}
int main(){
    initial();
    int t;
    scanf("%d",&t);
    for(int i=1; i<=t;i++){
        int n,x;
        scanf("%d",&n);
        long long ans=0;
        for(int j=0; j<n; j++){
            scanf("%d",&x);
            ans+=prime[lower_bound(prime,prime+cnt,x+1)-prime];
        }
        printf("Case %d: %lld Xukha\n",i,ans);
    }
    return 0;
}

[C - Aladdin and the Flying Carpet]

//shulun3
//解法： 
//主要利用公式: 
//一個整數ｎ可以表示爲若干素數乘積：　n = p1^a1 * p2^a2*…*pm^am; 
//則 n 的正因數的個數可以表示爲： num = (a1+1)*(a2+1)…(am+1);
//再減去1-min的
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<bitset>
#define CLR(a,v) memset(a,v,sizeof a)
using namespace std;
const int maxn = 1000005;
typedef long long ll;

bitset<maxn> vis;
int prime[maxn/10];
int p;
inline ll in()
{
    ll res=0;char c;
    while((c=getchar())<'0' || c>'9');
    while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
    return res;
}

void getPrime(){
    for(int i = 2; i< maxn; i++){
        if(!vis[i]){
            prime[p++]=i;
            for(int j = i+ i; j < maxn; j+=i){
                vis[j] = 1;
            }
        }
    }
}

int main(){
    getPrime();
    int T =in(),t=0;
    while(t++<T){
        ll area = in(),min = in();
        if(min>=sqrt(area)){
            printf("Case %d: %d\n",t,0);
            continue;
        }
        ll tmp = area;
        int ans = 1;
        for(int i = 0; 1LL*prime[i]*prime[i]<=area;i++){
            int cnt = 0;
            while(area%prime[i] == 0){
                cnt++;
                area/=prime[i];
            }
            ans*=(cnt+1);
        }
        if(area==1)ans>>=1;
        for(int i = 1; i< min; i++)if(tmp%i==0)ans--;
        printf("Case %d: %d\n",t,ans);
    }
    return 0;
}

[E - Leading and Trailing]

//http://www.th7.cn/Program/cp/201607/904075.shtml
//shulun5+快速模模版+log
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
typedef long long ll;
using namespace std;

int qmod(long long n , long long k){
    ll res = 1;
    while(k>0){
        if(k&1)res=(res*n)%1000;
        n = (n*n)% 1000;
        k >>= 1;
    }
    return res%1000;
}

int qlog(long long n, long long k){
    double a = log10(n+0.0)*k;
    if(a<=3)return pow(10,a); //本句神坑。。。。
    a-=(ll) a;
    int ans = pow(10.0,a)*1000;
    while(ans>=1000){ans/=10;}
    return ans;
}

int main(){
    int T,t=0;
    scanf("%d",&T);
    ll n,k;
    while(t++<T){
        scanf("%lld%lld",&n,&k);
        printf("Case %d: %03d %03d\n",t,qlog(n,k),qmod(n,k));
    }
    return 0;
}

[F - Goldbach`s Conjecture]

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<bitset>
#define maxn 10000005
using namespace std;
bitset<maxn> a;
int prime[maxn/10];
int cont=0;

inline void getPrime(){
    for(int i = 2; i< maxn; i++){
        if(!a[i]){
            prime[cont++] = i;
            for(int j = i+i; j < maxn ; j+=i)
                a[j] = 1;
        }
    }
}

int main(){
    getPrime();
    int T,t=0;
    scanf("%d",&T);
    int ans;
    while(t++<T){
        ans=0;
        int n;scanf("%d",&n);
        for(int i = 0;prime[i]<=n/2;i++){
            if(!a[n-prime[i]])ans++;
        }
        printf("Case %d: %d\n",t,ans);
    }
    return 0;
}

[G - Harmonic Number (II)]

#include<cstdio>
#include<cmath>

using namespace std;

int main(){
    int t=0,T;
    scanf("%d",&T);
    while(t++<T){
        long long n;
        scanf("%lld",&n);
        int m = sqrt(n);
        long long ans = 0;
        for(int i = 1; i <=m; i++){
            ans+=n/i;//oneprat
            ans+=i*(n/i-n/(i+1));//anotheraprt
        }
        if(n/m==m)ans-=n/m;//ji
        printf("Case %d: %lld\n",t,ans);
    }
    return 0;
}

[H - Pairs Forming LCM]

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<bitset>
#define maxn 10000000
using namespace std;
bitset<maxn> a;
int prime[maxn/10];
int cont=0;

inline void getPrime(){
    for(int i = 2; i< maxn; i++){
        if(!a[i]){
            prime[cont++] = i;
            for(int j = i+i; j < maxn ; j+=i)
                a[j] = 1;
        }
    }
}
int main(){
    getPrime();
    int T,t=0;
    scanf("%d",&T);
    while(t++<T){
        long long n,ans = 1,m;
        scanf("%lld",&n);
        m = n;
        for(int i = 0; prime[i] <= m/prime[i] && i < cont-1; ++i){
            int cot = 0;
            while(m%prime[i]==0){
                cot++;
                m/=prime[i];
            }
            ans*=(cot*2+1);
        }
        if(m>1)ans*=3;
        ans = (ans+1)/2;
        printf("Case %d: %lld\n",t,ans);
    }
    return 0;
}

[I - Harmonic Number]

//正常做法：技巧打表
#include<cstdio>
#define maxn  1e8
using namespace std;
double a[(int)maxn/50+10];

void init(){
    a[0] = 0;
    for(int i = 0; i <= maxn/50+9; i++){
        double tmp = 0;
        for(int j = i*50+1; j <= i*50+50;j++){
            tmp+=1.0/j;
        }
        a[i+1] = a[i]+tmp;
    }
}

int main(){
    init();
    int t=0,T;
    scanf("%d",&T);
    while(t++<T){
        double ans = 0;
        int n;scanf("%d",&n);
        ans+=a[n/50];
        for(int i = n/50*50+1; i <= n; i++){
            ans+=1.0/i;
        }
        printf("Case %d: %.10f\n",t,ans);
    }
    return 0;
}
//數學大佬的做法
#include<cstdio>
#include<cmath>
//oulachangshu:0.57721566490153286060651209
using namespace std;
double a[10010];
double r=0.57721566490153;
int main(){
    int n,i,cas=0,nn;
    for(int i=1;i<=10000;i++)
        a[i]=1.0/i+a[i-1];
    scanf("%d",&nn);
    while(cas++<nn){
        scanf("%d",&n);
        double i,ans=0;
        printf("Case %d: %.10f\n",cas,n>10000?1.0/(2*n)+log(n)+r:a[n]);
    }
}

[J - Mysterious Bacteria]

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<bitset>
#define maxn 70000
#define CLR(a,v) memset(a,v,sizeof a)
using namespace std;
bitset<maxn> a;
int prime[maxn];
int cont=0;
int vis[maxn]={0};
inline void getPrime(){
    for(int i = 2; i< maxn; i++){
        if(!a[i]){
            prime[cont++] = i;
            for(int j = i+i; j < maxn ; j+=i)
                a[j] = 1;
        }
    }
}
int main(){
    getPrime();
    int T,t=0;
    scanf("%d",&T);
    while(t++<T){
        CLR(vis,0);
        long long n,ans = 1,m;
        scanf("%lld",&n);
        int flag2= 0;
        if(n<0){n=-n;flag2= 1;}
        m = n;
        int minx = 1000;
        int lu = 0;
        for(int i = 0; prime[i] <= m/prime[i] && i < cont-1; ++i){
            if(m%prime[i]!=0){continue;}
            while(m%prime[i]==0){
                m/=prime[i];
                vis[i]++;
            }
            lu = i;
            minx=min(vis[i],minx);
        }
        if(m>1){printf("Case %d: 1\n",t);continue;}
        int flag = 0;
        for(int i = 0; i<=lu;i++){
            if(vis[i]==0)continue;
            if(vis[i]%minx!=0){
                flag=1;
                break;
            }
        }
        if(flag)minx=1;
        if(flag2){
            while(minx%2==0){
                minx/=2;
            }
        }
        printf("Case %d: %lld\n",t,minx);
    }
    return 0;
}

[K - Large Division]

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>

using namespace std;
char input[300];
long long b;
int main(){
    int t=0,T;
    scanf("%d",&T);
    while(t++<T){
        scanf("%s%lld",input,&b);
        if(b<0)b=-b;
        long long ans=0;
        int i=0;
        for(;input[i]<'0' || input[i]>'9';i++);
        for(;input[i]<='9' && input[i]>='0';i++){
            ans=ans*10+(input[i]-'0');
            ans%=b;
        }
        printf("Case %d: %sdivisible\n",t,(ans==0)?"":"not ");
    }
    return 0;
}

[L - Fantasy of a Summation]

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int mod,temp,n,k;
long long qpow(int n, int k){
    long long ans=1;
    while(k){
        if(k&1){ans=ans*n%mod;}
        n = n*n %mod;
        k>>=1;
    }
    return ans%mod;
}

int main(){
    int t=0,T;
    scanf("%d",&T);
    while(t++<T){
        long long sum=0;
        scanf("%d%d%d",&n,&k,&mod);
        for(int i = 0; i<n ; i++){
            scanf("%d",&temp);
            sum+=temp;
            sum%=mod;//漏了這句就wa了
        }
        sum*=k;
        printf("Case %d: %lld\n",t,(qpow(n,k-1)*sum)%mod);
    }
    return 0;
}

[M - Help Hanzo]

#include<cstdio>
using namespace std;
int pd[10000];//存放質數
bool judge[50010] = {0};//判斷50000內的質數
void init(){
    for(int i = 2; i <= 500; i++){
        if(!judge[i]){
            for(int j = i*i; j <= 50000; j+=i){
                judge[j] = 1;
            }
        }
    }
    int j = 0;
    for(int i = 2; i<= 50000; i++){
        if(!judge[i])pd[j++]=i;
    }
}

long long get(long long a, long long b){
    long long sum = 0;//計數
    bool sspd[200005]={0};
    if(a<2){a=2;}//1不是
    for(int i = 0; pd[i] <= b/pd[i]; i++){
        long long j = pd[i]*(a/pd[i]);
        if(j<a)j+=pd[i];
        if(j==pd[i])j+=(long long)pd[i];//不標記質數
        for(;j<=b; j+=pd[i])sspd[j-a]=1;//標記質數
    }
    for(int i = 0; i < b-a+1; i++){
        if(!sspd[i])sum++;
    }
    return sum;
}

int main(){
    init();
    int t=0,T,a,b;
    scanf("%d",&T);
    while(t++<T){
        scanf("%d%d",&a,&b);
        printf("Case %d: %lld\n",t,get(a,b));
    }
    return 0;
}

[N - Trailing Zeroes (III)]

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<bitset>
using namespace std;

int Find(int x){
    int sum = 0;
    while(x/5){
        sum+=x/5;
        x/=5;
    }
    return sum;
}


int main(){
    int t=0,T,a;
    scanf("%d",&T);
    while(t++<T){
        scanf("%d",&a);
        int b = a*4/5*5;
        while(Find(b)<a){
            b+=5;
        }
        if(Find(b) != a)printf("Case %d: impossible\n",t);
        else printf("Case %d: %d\n",t,b);
    }
    return 0;
}
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