Codeforces Round #Pi (Div. 2) (A,B,C,D)

Question A:
Title Address: Lineland Mail

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>Using namespace Std;typedef long Long ll;constintinf=0x3f3f3f3f; const double pi= ACOs (-1.0); const Double esp=1E-6; constintmaxn=1e5+Ten; struct node{intIdintValueintMax;intMin;}Q[MAXN];intCMP (struct node a,struct node B) {returnA.value<b.value;}intCMP1 (struct node a,struct node B) {returnA.id<b.id;}intMain () {intN,i; while(~SCANF ("%d", &n)) { for(i=1; i<=n;i++) {scanf ("%d",&Q[i]. value);Q[i]. id=i;Q[q[i]. id]. max=0;Q[q[i]. id]. min=0; }Sort(Q+1,Q+n+1, CMP); for(i=1; i<=n;i++) {if(i==1){Q[i]. min=ABS(Q[1]. value-Q[2]. value);Q[i]. max=ABS(Q[n]. value-Q[1]. value); }Else if(i==n) {Q[i]. min=ABS(Q[n]. value-Q[n-1]. value);Q[i]. max=ABS(Q[n]. value-Q[1]. value); }Else{Q[i]. Min=min (ABS(Q[i]. value-Q[i-1]. Value),ABS(Q[i+1]. value-Q[i]. value));Q[i]. Max=max (ABS(Q[i]. value-Q[1]. Value),ABS(Q[n]. value-Q[i]. value)); }        }Sort(Q+1,Q+1+N,CMP); for(i=1; i<=n;i++) {printf("%d %d\ n",Q[i]. Min,Q[i].        MAX); }    }return 0;}

Question B:
Title Address: Berland National Library

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>using namespace STD;typedef Long LongLL;Const intinf=0x3f3f3f3f;Const DoublePi=ACOs(-1.0);Const Doubleesp=1e-6;Const intmaxn= the;Const intmaxn1=1e6+Ten;Charstr[Ten];intAintVIS[MAXN1];intMain () {intNintCnt1,cnt2; while(~scanf("%d", &n)) {memset(Vis,0,sizeof(VIS)); Cnt1=cnt2=0; for(intI=0; i<n;i++) {scanf('%s%d ', &str,&a);if(str[0]==' + ') {cnt1++; vis[a]=1;                Cnt2=max (Cnt1,cnt2); }Else{if(!vis[a])                    {cnt2++; }Else{vis[a]=0;                    cnt1--; }                }            }printf("%d\n", Cnt2); }return 0;}

Question C:
Title Address: Geometric progression
Test instructions: The number of geometric series of a sequence in which the k is the male ratio and the number of items is 3.
Idea: Here I used a map to open a dp[i][j] array, record the length of I with the number of J-ending numbers, so we can easily draw dp[i][j]+=dp[i-1][j/k],dp[1][j]=1. Because I use X%K maintenance in the program, so I want to push backwards. Because the data problem map does not open map

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>using namespace STD;typedef Long LongLL;Const intinf=0x3f3f3f3f;Const DoublePi=ACOs(-1.0);Const Doubleesp=1e-6;Const intmaxn=200010; LL A[MAXN];intMain () { map<LL,LL>dp[4];intN,k,i; LL ans; while(~scanf("%d%d", &n,&k)) {ans=0; for(i=0;i<4; i++) dp[i].clear (); for(i=1; i<=n;i++)scanf("%lld", &a[i]); for(i=1; i<=n;i++) {if(a[i]%k==0) {dp[3][a[i]]+=dp[2][A[I]/K]; dp[2][a[i]]+=dp[1][A[I]/K]; } dp[1][a[i]]++; } map<LL,LL>:: Iterator it; for(it=dp[3].begin (); it!=dp[3].end (); it++) ans+=it->second;printf("%lld\n", ans); }return 0;}

Question d:
Title Address: one-dimensional Battle Ships
Test instructions: Given an interval, each time a point is removed, it is required to be put down in the remaining blanks, K is a range (a interval cannot be absent). If it can be put down, then output-1, otherwise the output in the place of the first few points can not be put down.
Idea: the interval [l,r] removes one point x at a time, and the remaining interval is [l,x-1],[x+1,r]. Then find the number of ships that can be lowered for each interval: (interval length + 1)/(length of the boat + 1) {Because the ship is absent from the ship, so divided by the length of the boat is +1}. Two-point answer to find the first few points let the number of ships less than K.

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>using namespace STD;typedef Long LongLL;Const intinf=0x3f3f3f3f;Const DoublePi=ACOs(-1.0);Const Doubleesp=1e-6;Const intmaxn=200010;intX[MAXN];intX1[MAXN];intN,k,a,m;intCntintFind (intm) { for(intI=1; i<=m;i++) X1[i]=x[i]; Sort (x1+1, x1+m+1);intlen=0; Cnt=0; for(intI=1; i<=m;i++) {cnt+= (X1[i]-len)/(A +1);    Len=x1[i]; } cnt+= (n+1-len)/(A +1);if(cnt>=k)return 1;Else        return 0;}intMain () {scanf(" %d%d%d", &n,&k,&a);scanf("%d", &m); for(intI=1; i<=m;i++)scanf("%d", &x[i]);intL=1, r=m;intMidintRes=inf; while(L&LT;=R) {mid= (l+r) >>1;if(Find (mid)) l=mid+1;Else{r=mid-1;        Res=min (Res,mid); }    }//printf ("mid==%d\n", mid);    //printf ("l==%d\n", L);    if(l>m)puts("-1");Else        printf("%d\n", res);}

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Codeforces Round #Pi (Div. 2) (A,B,C,D)