contest 1.14

B.数论number

欧拉降幂定理

#include <iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
typedef long long ll;
using namespace std;
 
ll a,b,c,p;
 
ll qpow(ll a,ll b,ll mod){
  ll ret=1;
  while(b){
    if(b&1) ret=ret*a%mod;
    a=a*a%mod;
    b>>=1;
  }
  return ret;
}
 
ll getphi(ll x){
    ll ret=x;
    for(ll i=2;i*i<=x;i++)
      if(x%i==0){
        ret/=i,ret*=(i-1);
        while(x%i==0)x/=i;
      }
    if(x>1)ret/=x,ret*=x-1;
    return ret;
}
 
 
int main()
{
    scanf("%lld%lld%lld%lld",&a,&b,&c,&p);
    ll phi=getphi(p);
    ll phi_=getphi(phi);
    ll tmp=qpow(b,c%phi_+phi_,phi);
    ll ans=qpow(a,tmp+phi,p);
    printf("%lld\n",ans);
    return 0;
}
View Code

 

C.约数的平方和

考虑每个约数对答案的贡献

#include <iostream>
#include<cstdio>
#include<cmath>
typedef long long ll;
using namespace std;
 
ll n,p;
 
int main()
{
    scanf("%lld%lld",&n,&p);
    ll ans=0;
    for(ll i=1;i<=1000005;i++){
        ans=(ans+i*i%p*(n/i)%p)%p;
    }
    printf("%lld\n",ans);
    return 0;
}
View Code

 

D.随机配对

 

H.幻想机器人

模拟

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<algorithm>
typedef long long ll;
using namespace std;
 
char s[100005];
 
struct Node{
  ll x,y;
}now;
 
map<pair<ll,ll>,ll> mp;
 
int main()
{
    ll n,m,k,len,x,y;
    scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&k,&len,&x,&y);
    scanf("%s",s);
    ll cnt=0;
    mp.clear();
    now.x=x,now.y=y;
    mp[make_pair(now.x,now.y)]=1,cnt++;
    for(ll i=0;i<len;i++){
        if(s[i]=='U'){
            if(now.x-1>=1){
                now.x--;
                if((i+1)%k==0){
                    if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n");
                    else mp[make_pair(now.x,now.y)]=1,cnt++;
                }
            }
            else{
                printf("AWaDa!\n");
                if((i+1)%k==0){
                    if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n");
                    else mp[make_pair(now.x,now.y)]=1,cnt++;
                }
            }
        }
        else if(s[i]=='D'){
            if(now.x+1<=n){
                now.x++;
                if((i+1)%k==0){
                    if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n");
                    else mp[make_pair(now.x,now.y)]=1,cnt++;
                }
            }
            else{
                printf("AWaDa!\n");
                if((i+1)%k==0){
                    if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n");
                    else mp[make_pair(now.x,now.y)]=1,cnt++;
                }
            }
        }
        else if(s[i]=='L'){
            if(now.y-1>=1){
                now.y--;
                if((i+1)%k==0){
                    if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n");
                    else mp[make_pair(now.x,now.y)]=1,cnt++;
                }
            }
            else{
                printf("AWaDa!\n");
                if((i+1)%k==0){
                    if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n");
                    else mp[make_pair(now.x,now.y)]=1,cnt++;
                }
            }
        }
        if(s[i]=='R'){
            if(now.y+1<=m){
                now.y++;
                if((i+1)%k==0){
                    if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n");
                    else mp[make_pair(now.x,now.y)]=1,cnt++;
                }
            }
            else{
                printf("AWaDa!\n");
                if((i+1)%k==0){
                    if(mp[make_pair(now.x,now.y)]) printf("AKTang!\n");
                    else mp[make_pair(now.x,now.y)]=1,cnt++;
                }
            }
        }
    }
    printf("%lld\n",n*m-cnt);
    return 0;
}
View Code

 

B.阿卡吃馅饼

大数+枚举

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<iomanip>
#include<algorithm>
using namespace std;
 
#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4
 
class BigNum
{
private:
    int a[1000];    //可以控制大数的位数
    int len;       //大数长度
public:
    BigNum(){ len = 1;memset(a,0,sizeof(a)); }   //构造函数
    BigNum(const int);       //将一个int类型的变量转化为大数
    BigNum(const char*);     //将一个字符串类型的变量转化为大数
    BigNum(const BigNum &);  //拷贝构造函数
    BigNum &operator=(const BigNum &);   //重载赋值运算符,大数之间进行赋值运算
 
    friend istream& operator>>(istream&,  BigNum&);   //重载输入运算符
    friend ostream& operator<<(ostream&,  BigNum&);   //重载输出运算符
 
    BigNum operator+(const BigNum &) const;   //重载加法运算符,两个大数之间的相加运算
    BigNum operator-(const BigNum &) const;   //重载减法运算符,两个大数之间的相减运算
    BigNum operator*(const BigNum &) const;   //重载乘法运算符,两个大数之间的相乘运算
    BigNum operator/(const int   &) const;    //重载除法运算符,大数对一个整数进行相除运算
 
    BigNum operator^(const int  &) const;    //大数的n次方运算
    int    operator%(const int  &) const;    //大数对一个int类型的变量进行取模运算
    bool   operator>(const BigNum & T)const;   //大数和另一个大数的大小比较
    bool   operator>(const int & t)const;      //大数和一个int类型的变量的大小比较
 
    void print();       //输出大数
};
BigNum::BigNum(const int b)     //将一个int类型的变量转化为大数
{
    int c,d = b;
    len = 0;
    memset(a,0,sizeof(a));
    while(d > MAXN)
    {
        c = d - (d / (MAXN + 1)) * (MAXN + 1);
        d = d / (MAXN + 1);
        a[len++] = c;
    }
    a[len++] = d;
}
BigNum::BigNum(const char*s)     //将一个字符串类型的变量转化为大数
{
    int t,k,index,l,i;
    memset(a,0,sizeof(a));
    l=strlen(s);
    len=l/DLEN;
    if(l%DLEN)
        len++;
    index=0;
    for(i=l-1;i>=0;i-=DLEN)
    {
        t=0;
        k=i-DLEN+1;
        if(k<0)
            k=0;
        for(int j=k;j<=i;j++)
            t=t*10+s[j]-'0';
        a[index++]=t;
    }
}
BigNum::BigNum(const BigNum & T) : len(T.len)  //拷贝构造函数
{
    int i;
    memset(a,0,sizeof(a));
    for(i = 0 ; i < len ; i++)
        a[i] = T.a[i];
}
BigNum & BigNum::operator=(const BigNum & n)   //重载赋值运算符,大数之间进行赋值运算
{
    int i;
    len = n.len;
    memset(a,0,sizeof(a));
    for(i = 0 ; i < len ; i++)
        a[i] = n.a[i];
    return *this;
}
istream& operator>>(istream & in,  BigNum & b)   //重载输入运算符
{
    char ch[MAXSIZE*4];
    int i = -1;
    in>>ch;
    int l=strlen(ch);
    int count=0,sum=0;
    for(i=l-1;i>=0;)
    {
        sum = 0;
        int t=1;
        for(int j=0;j<4&&i>=0;j++,i--,t*=10)
        {
            sum+=(ch[i]-'0')*t;
        }
        b.a[count]=sum;
        count++;
    }
    b.len =count++;
    return in;
 
}
ostream& operator<<(ostream& out,  BigNum& b)   //重载输出运算符
{
    int i;
    cout << b.a[b.len - 1];
    for(i = b.len - 2 ; i >= 0 ; i--)
    {
        cout.width(DLEN);
        cout.fill('0');
        cout << b.a[i];
    }
    return out;
}
 
BigNum BigNum::operator+(const BigNum & T) const   //两个大数之间的相加运算
{
    BigNum t(*this);
    int i,big;      //位数
    big = T.len > len ? T.len : len;
    for(i = 0 ; i < big ; i++)
    {
        t.a[i] +=T.a[i];
        if(t.a[i] > MAXN)
        {
            t.a[i + 1]++;
            t.a[i] -=MAXN+1;
        }
    }
    if(t.a[big] != 0)
        t.len = big + 1;
    else
        t.len = big;
    return t;
}
BigNum BigNum::operator-(const BigNum & T) const   //两个大数之间的相减运算
{
    int i,j,big;
    bool flag;
    BigNum t1,t2;
    if(*this>T)
    {
        t1=*this;
        t2=T;
        flag=0;
    }
    else
    {
        t1=T;
        t2=*this;
        flag=1;
    }
    big=t1.len;
    for(i = 0 ; i < big ; i++)
    {
        if(t1.a[i] < t2.a[i])
        {
            j = i + 1;
            while(t1.a[j] == 0)
                j++;
            t1.a[j--]--;
            while(j > i)
                t1.a[j--] += MAXN;
            t1.a[i] += MAXN + 1 - t2.a[i];
        }
        else
            t1.a[i] -= t2.a[i];
    }
    t1.len = big;
    while(t1.a[len - 1] == 0 && t1.len > 1)
    {
        t1.len--;
        big--;
    }
    if(flag)
        t1.a[big-1]=0-t1.a[big-1];
    return t1;
}
 
BigNum BigNum::operator*(const BigNum & T) const   //两个大数之间的相乘运算
{
    BigNum ret;
    int i,j,up;
    int temp,temp1;
    for(i = 0 ; i < len ; i++)
    {
        up = 0;
        for(j = 0 ; j < T.len ; j++)
        {
            temp = a[i] * T.a[j] + ret.a[i + j] + up;
            if(temp > MAXN)
            {
                temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
                up = temp / (MAXN + 1);
                ret.a[i + j] = temp1;
            }
            else
            {
                up = 0;
                ret.a[i + j] = temp;
            }
        }
        if(up != 0)
            ret.a[i + j] = up;
    }
    ret.len = i + j;
    while(ret.a[ret.len - 1] == 0 && ret.len > 1)
        ret.len--;
    return ret;
}
BigNum BigNum::operator/(const int & b) const   //大数对一个整数进行相除运算
{
    BigNum ret;
    int i,down = 0;
    for(i = len - 1 ; i >= 0 ; i--)
    {
        ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
        down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
    }
    ret.len = len;
    while(ret.a[ret.len - 1] == 0 && ret.len > 1)
        ret.len--;
    return ret;
}
int BigNum::operator %(const int & b) const    //大数对一个int类型的变量进行取模运算
{
    int i,d=0;
    for (i = len-1; i>=0; i--)
    {
        d = ((d * (MAXN+1))% b + a[i])% b;
    }
    return d;
}
BigNum BigNum::operator^(const int & n) const    //大数的n次方运算
{
    BigNum t,ret(1);
    int i;
    if(n<0)
        exit(-1);
    if(n==0)
        return 1;
    if(n==1)
        return *this;
    int m=n;
    while(m>1)
    {
        t=*this;
        for( i=1;i<<1<=m;i<<=1)
        {
            t=t*t;
        }
        m-=i;
        ret=ret*t;
        if(m==1)
            ret=ret*(*this);
    }
    return ret;
}
bool BigNum::operator>(const BigNum & T) const   //大数和另一个大数的大小比较
{
    int ln;
    if(len > T.len)
        return true;
    else if(len == T.len)
    {
        ln = len - 1;
        while(a[ln] == T.a[ln] && ln >= 0)
            ln--;
        if(ln >= 0 && a[ln] > T.a[ln])
            return true;
        else
            return false;
    }
    else
        return false;
}
bool BigNum::operator >(const int & t) const    //大数和一个int类型的变量的大小比较
{
    BigNum b(t);
    return *this>b;
}
 
void BigNum::print()    //输出大数
{
    int i;
    cout << a[len - 1];
    for(i = len - 2 ; i >= 0 ; i--)
    {
        cout.width(DLEN);
        cout.fill('0');
        cout << a[i];
    }
    cout << endl;
}
 
BigNum qpow(BigNum a,int b){
  BigNum ret=1;
  while(b){
    if(b&1) ret=ret*a;
    a=a*a;
    b>>=1;
  }
  return ret;
}
 
int main()
{
   int T;
   scanf("%d",&T);
   while(T--){
     int p,r;scanf("%d%d",&p,&r);
     int x;
     BigNum now(1),tot(0);
     for(x=0;x<=1000;x++){
        tot=tot+now*p;
        //tot.print();
        if(tot%r==0) break;
        now=now*10;
     }
     if(x>1000) puts("GG");
     else printf("%d\n",x);
   }
}
View Code

 

E.大战幻想珠

暴力

#include <iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;

char s[3005];

bool check(char a,char b){
  if(a==b||a=='?'||b=='?') return 1;
  return 0;
}

int main()
{
    int n;scanf("%d",&n);
    scanf("%s",s);
    int ans=0;
    for(int i=0;i<n;i++){
        for(int j=0;i-j>=0&&i+j<n;j++){
            if(check(s[i-j],s[i+j])) ans++;
            else break;
        }
        for(int j=0;i+j<n&&i-j-1>=0;j++){
            if(check(s[i-j-1],s[i+j])) ans++;
            else break;
        }
    }
    printf("%d\n",ans);
    return 0;
}
View Code

 

C.烤乐滋打虎

猜你喜欢

转载自www.cnblogs.com/lllxq/p/10269099.html
0条评论
添加一条新回复