In this HackerEarth New Language problem solution A new mathematical language (let's call it Ninja) is discovered, in which 8 is completely missing.
So, new counting becomes (1,2,3,4,5,6,7,9,10,...). Whole counting is written without 8.
For example, 29 comes after 27,90 comes after 79, etc.
Thus 27 + 1 = 29, 79 + 1 = 90, .
Now, your task is to perform addition, subtraction, multiplication, and division operations on two numbers in this new language Ninja.
HackerEarth New Language problem solution.
#include<bits/stdc++.h>
typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;
typedef double lf;
typedef long double llf;
typedef std::pair<int,int> pii;
#define xx first
#define yy second
template<typename T> inline T max(T a,T b){return a>b?a:b;}
template<typename T> inline T min(T a,T b){return a<b?a:b;}
template<typename T> inline T abs(T a){return a>0?a:-a;}
template<typename T> inline bool repr(T &a,T b){return a<b?a=b,1:0;}
template<typename T> inline bool repl(T &a,T b){return a>b?a=b,1:0;}
template<typename T> inline T gcd(T a,T b){T t;if(a<b){while(a){t=a;a=b%a;b=t;}return b;}else{while(b){t=b;b=a%b;a=t;}return a;}}
template<typename T> inline T sqr(T x){return x*x;}
#define mp(a,b) std::make_pair(a,b)
#define pb push_back
#define I inline
#define mset(a,b) memset(a,b,sizeof(a))
#define mcpy(a,b) memcpy(a,b,sizeof(a))
#define fo0(i,n) for(int i=0,i##end=n;i<i##end;i++)
#define fo1(i,n) for(int i=1,i##end=n;i<=i##end;i++)
#define fo(i,a,b) for(int i=a,i##end=b;i<=i##end;i++)
#define fd0(i,n) for(int i=(n)-1;~i;i--)
#define fd1(i,n) for(int i=n;i;i--)
#define fd(i,a,b) for(int i=a,i##end=b;i>=i##end;i--)
#define foe(i,x)for(__typeof((x).end())i=(x).begin();i!=(x).end();++i)
#define fre(i,x)for(__typeof((x).rend())i=(x).rbegin();i!=(x).rend();++i)
struct Cg{I char operator()(){return getchar();}};
struct Cp{I void operator()(char x){putchar(x);}};
#define OP operator
#define RT return *this;
#define UC unsigned char
#define RX x=0;UC t=P();while((t<'0'||t>'9')&&t!='-')t=P();bool f=0;\
if(t=='-')t=P(),f=1;x=t-'0';for(t=P();t>='0'&&t<='9';t=P())x=x*10+t-'0'
#define RL if(t=='.'){lf u=0.1;for(t=P();t>='0'&&t<='9';t=P(),u*=0.1)x+=u*(t-'0');}if(f)x=-x
#define RU x=0;UC t=P();while(t<'0'||t>'9')t=P();x=t-'0';for(t=P();t>='0'&&t<='9';t=P())x=x*10+t-'0'
#define TR *this,x;return x;
I bool IS(char x){return x==10||x==13||x==' ';}template<typename T>struct Fr{T P;I Fr&OP,(int&x)
{RX;if(f)x=-x;RT}I OP int(){int x;TR}I Fr&OP,(ll &x){RX;if(f)x=-x;RT}I OP ll(){ll x;TR}I Fr&OP,(char&x)
{for(x=P();IS(x);x=P());RT}I OP char(){char x;TR}I Fr&OP,(char*x){char t=P();for(;IS(t);t=P());if(~t){for(;!IS
(t)&&~t;t=P())*x++=t;}*x++=0;RT}I Fr&OP,(lf&x){RX;RL;RT}I OP lf(){lf x;TR}I Fr&OP,(llf&x){RX;RL;RT}I OP llf()
{llf x;TR}I Fr&OP,(uint&x){RU;RT}I OP uint(){uint x;TR}I Fr&OP,(ull&x){RU;RT}I OP ull(){ull x;TR}};Fr<Cg>in;
#define WI(S) if(x){if(x<0)P('-'),x=-x;UC s[S],c=0;while(x)s[c++]=x%10+'0',x/=10;while(c--)P(s[c]);}else P('0')
#define WL if(y){lf t=0.5;for(int i=y;i--;)t*=0.1;if(x>=0)x+=t;else x-=t,P('-');*this,(ll)(abs(x));P('.');if(x<0)\
x=-x;while(y--){x*=10;x-=floor(x*0.1)*10;P(((int)x)%10+'0');}}else if(x>=0)*this,(ll)(x+0.5);else *this,(ll)(x-0.5);
#define WU(S) if(x){UC s[S],c=0;while(x)s[c++]=x%10+'0',x/=10;while(c--)P(s[c]);}else P('0')
template<typename T>struct Fw{T P;I Fw&OP,(int x){WI(10);RT}I Fw&OP()(int x){WI(10);RT}I Fw&OP,(uint x){WU(10);RT}
I Fw&OP()(uint x){WU(10);RT}I Fw&OP,(ll x){WI(19);RT}I Fw&OP()(ll x){WI(19);RT}I Fw&OP,(ull x){WU(20);RT}I Fw&OP()
(ull x){WU(20);RT}I Fw&OP,(char x){P(x);RT}I Fw&OP()(char x){P(x);RT}I Fw&OP,(const char*x){while(*x)P(*x++);RT}
I Fw&OP()(const char*x){while(*x)P(*x++);RT}I Fw&OP()(lf x,int y){WL;RT}I Fw&OP()(llf x,int y){WL;RT}};Fw<Cp>out;
const int N=1005;
int S,X,n;
pii s[N];
int main()
{
in,S,X,n;
fo1(i,n)in,s[i].xx,s[i].yy;
std::sort(s+1,s+n+1);
int lst=0;
fo1(i,n)
{
if(ll(s[i].xx-1-lst)*X>=S)
{
out,lst+(S+X-1)/X,'\n';
return 0;
}
S-=ll(s[i].xx-1-lst)*X;
if(S<=s[i].yy)
{
out,s[i].xx,'\n';
return 0;
}
S-=s[i].yy;
lst=s[i].xx;
}
out,lst+(S+X-1)/X,'\n';
}#include<bits/stdc++.h>
using namespace std;
#define N 100005
#define ll long long
#define pb push_back
#define mp make_pair
#define fi first
#define se second
//Mistakes:
#define pritnf printf
template <typename T>
string NumberToString ( T Number )
{
stringstream ss;
ss << Number;
return ss.str();
}
ll StringToNumber (string str)
{
stringstream sstr(str);
ll val;
sstr >> val;
return val;
}
ll base9(ll n)
{
ll ret=0;
ll p=1;
while(n!=0)
{
ret=(n%9)*p+ret;
n/=9;
p*=10;
}
return ret;
}
ll revBase9(ll n)
{
ll ret=0,p=0;
while(n!=0)
{
ret+=(n%10)*pow(9,p);
p++;
n/=10;
}
return ret;
}
ll fromNinja(ll n)
{
string s=NumberToString(n);
for(int i=0;i<s.length();i++)
if(s[i]=='9')
s[i]='8';
n=StringToNumber(s);
return revBase9(n);
}
ll toNinja(ll n)
{
n=base9(n);
string s=NumberToString(n);
for(int i=0;i<s.length();i++)
if(s[i]=='8')
s[i]='9';
return StringToNumber(s);
}
int main()
{
ll a[N],t,i=0,j,k,x,y,z,count=0,p,flag=0,ans=0,sum=0,l,n,m,max1,min1,pos,tmp,q;
char c;
scanf("%lld",&t);
while(t--)
{
scanf("%lld%c%lld",&p,&c,&q);
//printf("%lld %c %lld\n",p,c,q);
p=fromNinja(p);
q=fromNinja(q);
//printf("%lld %lld\n",p,q);
switch(c)
{
case '+':
ans=p+q;
break;
case '-':
ans=p-q;
break;
case '*':
ans=p*q;
break;
case '/':
ans=p/q;
break;
}
printf("%lld\n",toNinja(ans));
}
return 0;
}Second solution
#include<bits/stdc++.h>
#define ll long long
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
string s,a="",b="";char op;
cin>>s;
for(int i=0;i<s.size();i++)
{
if(s[i]=='+' || s[i]=='-' || s[i]=='*' || s[i]=='/')
{
op=s[i];
a=b;
b="";
continue;
}
b+=(s[i]=='9')?'8':s[i];
}
ll vala=0,valb=0,mul=1;
for(int i=a.size()-1;i>=0;i--)
{
vala=(a[i]-48)*mul+vala;
mul*=9;
}
mul=1;
for(int i=b.size()-1;i>=0;i--)
{
valb=(b[i]-48)*mul+valb;
mul*=9;
}
ll ans;
if(op=='+')ans=vala+valb;
else if(op=='-')ans=vala-valb;
else if(op=='*')ans=vala*valb;
else ans=vala/valb;
a="";
while(ans)
{
char temp=(char)(ans%9+'0');
if(temp=='8')temp='9';
a=temp+a;
ans/=9;
}
if(a=="")a+='0';
cout<<a<<"\n";
}
return 0;
}
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