## contest 1.14

B.数论number

```#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

```#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

```#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{
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{
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{
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{
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

```#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

```#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

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