Codeforces Round #489 (Div. 2)

codeforces Round #489 (Div. 2)A. Nastya and an Array

#include <bits/stdc++.h>#define REP (i,a,b) for (int i=a;i<=b;++i)#define FREP (I,A,B) for (int i=a;i>=b;--i)#define MEM (w) memset (w,0,sizeof (w) )#define PB Push_backtypedef Long LongllConst intN = -;Const intINF =0x3f3f3f3f;inline intRead () {CharC=getchar ();intx=0, f=1; while(c<' 0 '|| C>' 9 '){if(c=='-') f=-1; C=getchar ();} while(c>=' 0 '&&c<=' 9 ') {x=x*Ten+c-' 0 '; C=getchar ();}returnX*f;}using namespaceStdintn,ans;map<int,int> M;intMain () {n = read (); Rep (I,1, N) {intX=read ();if(x&&! M[x]) {m[x]=1;        ++ans; }} printf ("%d\n", ans);return 0;}

B. Nastya Studies Informatics\ (GCD (A, B) = x, LCM (A, b) = y\), i.e.\ (gcd (\frac {a}{x},\frac {b}{x}) = 1, \frac {a*b}{x} = \frac{a}{x}*b = y\), then we can enumerate directly\ (\frac{a}{x}\)To check if a, B is an answer within [l,r], just enumerate\ (\sqrt{y}\)The number can be within. (the card too long led to c no time to check ...) It's just cold.

#include <bits/stdc++.h>#define REP (i,a,b) for (int i=a;i<=b;++i)#define FREP (I,A,B) for (int i=a;i>=b;--i)#define MEM (w) memset (w,0,sizeof (w) )#define PB Push_backtypedef Long LongllConst intN = -;Const intINF =0x3f3f3f3f;inlinell read () {CharC=getchar (); ll x=0, f=1; while(c<' 0 '|| C>' 9 '){if(c=='-') f=-1; C=getchar ();} while(c>=' 0 '&&c<=' 9 ') {x=x*Ten+c-' 0 '; C=getchar ();}returnX*f;}using namespaceStd;ll x,y,l,r,ans,ans2;set< pair<ll,ll> > S;ll LCM (ll a,ll b) {returnA*B/__GCD (A, b);}intMain () {l=read (); R=read (); X=read (); Y=read ();if(x==y) {if(L&LT;=X&AMP;&AMP;X&LT;=R)returnPuts"1"),0;Else returnPuts"0"),0; }if(y%x| | Y&LT;X)returnPuts"0"),0;if(y==1){if(x==1&AMP;&AMP;L&LT;=X&AMP;&AMP;X&LT;=R)returnPuts"1"),0;Else returnPuts"0"),0; } for(LL i=1; i*i<=y;++i) {if(y% (x*i) = =0&AMP;&AMP;__GCD (i*x,y/i) ==X&AMP;&AMP;LCM (i*x,y/i) ==y) {if(l<=i*x&&y/i<=r&&l<=y/i&&i*x<=r) S.insert (Make_pair (i*x,y/i)), S.insert (make        _pair (y/i,i*x)); }} printf ("%d\n",(int) s.size ());return 0;}

C. Nastya and a wardrobe slightly expanded to write, you can find the law, and then the formula is very obvious, see the code specifically. Pay attention to the Special award.

#include <bits/stdc++.h>#define REP (i,a,b) for (int i=a;i<=b;++i)#define FREP (I,A,B) for (int i=a;i>=b;--i)#define MEM (w) memset (w,0,sizeof (w) )#define PB Push_backtypedef Long LongllConst intN = -;Const intINF =0x3f3f3f3f;inline intRead () {CharC=getchar ();intx=0, f=1; while(c<' 0 '|| C>' 9 '){if(c=='-') f=-1; C=getchar ();} while(c>=' 0 '&&c<=' 9 ') {x=x*Ten+c-' 0 '; C=getchar ();}returnX*f;}using namespaceStdConst intMoD =1e9+7; ll Q_pow (ll A,ll b) {ll ans=1; while(b) {if(b&1) ans= (ans*a)%mod;        A= (a*a)%mod; b>>=1; }returnAns;} ll N,k;intMain () {scanf ("%i64d%i64d", &n,&k);if(n==0)returnPuts"0"),0;    N%=mod; printf"%i64d\n", ((Q_pow (2K+1) *n)%mod-(Q_pow (2K-1)%mod+mod)%mod);return 0;}

D. Nastya and a Game\ (p = s*k≤2*10^{18}\), so if each multiply by a number greater than 1, p requires at most\ (\log{(2*10^{18})}\)multiplied by the number. Then just press test instructions to enumerate the left endpoint, each time jumping to the next non-1 position, for a continuous period of 1, calculate the maximum, the minimum value can be judged whether to include the location of the condition.

#include <bits/stdc++.h>typedef Long LongllConst intN =2e5+ -;using namespaceStdintN,k;ll A[n],sum[n];intNxt[n];voidInit () { for(intI=1; i<=n;++i) Sum[i]=sum[i-1]+a[i]; Nxt[n] = n+1; for(intI=n-1; i>=1;-I.) {if(A[i+1] ==1) Nxt[i] = Nxt[i+1];ElseNxt[i] = i+1; }}intCk1 (ll A, ll b) {returnA *1.0> long_max*1.0/b;}intCK (ll x1, ll x2, ll p) {if(p%k==0) {p/=k;if(X1&LT;=P&AMP;&AMP;P&LT;=X2)return 1; }return 0;}intMain () {scanf ("%d%d", &n,&k); for(intI=1; i<=n;++i) scanf ("%i64d", &a[i]); Init ();intAns =0; for(intL=1; L<=n;++l) {ll x=1; for(intR=l; r<=n; R=nxt[r]) {if(Ck1 (x, A[r])) Break; X*=A[R];if(CK (sum[r]-sum[l-1], Sum[nxt[r]-1]-sum[l-1], x)) ++ans; }} printf ("%d\n", ans);}

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Codeforces Round #489 (Div. 2)