HackerEarth City Travel problem solution YASH PAL, 31 July 2024 In this HackerEarth City Travel problem solution, You are going from City A to City B. The distance between A and B is S km. On most days, you can go at most X km one day. But there are N exceptions, on the Ti the day, you can go at most Yi km. You need to find out the minimum number of days required to reach city B from city A. HackerEarth City Travel 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 secondtemplate<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';} coding problems