B. hxc write
AC code:
#pragma GCC optimize(2) #include <cstdio> #include <queue> #include <string> #include <cstring> #include <algorithm> #include <cstdlib> #include <iostream> #include <iomanip> #include <cmath> #include <vector> #include <set> #include <map> #include <fstream> #include <cassert> #define ll long long #define R register int #define I inline void #define lc c[x][0] #define rc c[x][1] using namespace std; const int INF = 0x3f3f3f3f; const int maxn = 1e6 + 10; int n,q; inline int read() { int x=0,f=0; char ch=0; while(!isdigit(ch)) {f|=ch=='-';ch=getchar();} while(isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=getchar(); return f?-x:x; } struct node { int num; int typ; int key; int ans; }pp[maxn]; vector<int> v; int getid(int x) { return lower_bound(v.begin(),v.end(),x) - v.begin(); } int con[maxn]; int nxt[maxn]; int cmp(node a,node b) { if(a.key != b.key) return a.key > b.key; return a.num < b.num; } int cmp1(node a,node b) { return a.num < b.num; } int find(int x) { if(x == nxt[x]) return x; return nxt[x] = find(nxt[x]); } int u[maxn]; int main() { n = read(); q = read(); for(int i = 1; i <= q; i++) { //pp[i].num = i; int a,b; a = read(); b = read(); pp[i].key = b; pp[i].typ = a; v.push_back(b); } sort(v.begin(),v.end()); v.erase(unique(v.begin(),v.end()),v.end()); int len = v.size(); //printf("len%d\n",len); for(int i = 1; i <= len; i++) { nxt[i] = i; //printf("qwe%d\n",v[i - 1]); } con[len - 1] = 0; for(int i = 0; i < len - 1; i++) { if(v[i] + 1 == v[i + 1]) con[i + 1] = 1; //printf("con%d\n",con[i]); } /*for(int i = 1; i <= len; i++) printf("%d %d\n",v[i - 1],con[i]); printf("\n");*/ for(int i = 1; i <= q; i++) { int key = getid(pp[i].key) + 1; //printf("key%d %d\n",pp[i].key,key); if(pp[i].typ == 1) { if(con[key]) { nxt[find(key)] = find(key + 1); } else { int temp = find(key); //printf("qwe%d\n",find(key)); u[find(key)] = 1; //printf("qwe%d\n",find(key)); } } else { if(u[find(key)] == 1) { //printf("f%d\n",nxt[key]); if(v[find(key) - 1] + 1 <= n) //pp[i].ans = v[find(key)] + 1; printf("%d\n",v[find(key) - 1] + 1); else //pp[i].ans = -1; printf("-1\n"); } else { //printf("124252\n"); printf("%d\n",v[find(key) - 1]); } } } /*for(int i = 1; i <= q; i++) if(pp[i].typ == 2) { printf("%d\n",pp[i].ans); }*/ }
C. Buy Watermelon
The meaning of problems: the volume w watermelon cut in half, the two halves of the weights are guaranteed even.
Ideas: water problems, ambiguous meaning of the questions.
AC code:
#include<cstdio> #include<algorithm> using namespace std; int w; int main(){ scanf("%d",&w); if(w==2||w%2==1) printf("NO\n"); else printf("YES\n"); return 0; }
D. Carnegion（kmp）
Meaning of the questions: Simplification means the problem to determine whether the string.
Thinking: kmp find substring, the time complexity of O (q * (S + T)).
AC code:
#include<cstdio> #include<algorithm> #include<cstring> #include<assert.h> using namespace std; static int * GetNext(const char *sub) { int lensub = strlen(sub); int *next = (int *)malloc(lensub*sizeof(int)); assert(next != NULL); next[0] = -1; next[1] = 0; int j = 1; int k = 0; while(j+1<lensub) { if((k==-1) || (sub[k]==sub[j])) next[++j] = ++k; else k = next[k]; } return next; } int KMP(const char *str,const char *sub,int pos) { assert(str!=NULL && sub!=NULL); int lenstr = strlen(str); int lensub = strlen(sub); if(pos<0 || pos>=lenstr) return -1; int *next = GetNext(sub); int i = pos; int j = 0; while(i<lenstr && j<lensub) { if(j==-1 || (str[i]==sub[j])) i++,j++; else j = next[j]; } free(next); if(j >= lensub) return i-j; else return -1; } const int maxn=1e5+5; int m,len1,len2; char T[maxn],S[maxn]; int main(){ scanf("%s",T); len1=strlen(T); scanf("%d",&m); while(m--){ scanf("%s",S); len2=strlen(S); if(len1>len2){ if(KMP(T,S,0)>=0) printf("my child!\n"); else printf("oh, child!\n"); } else if(len1==len2){ if(strcmp(T,S)==0) printf("jntm!\n"); else printf("friend!\n"); } else{ if(KMP(S,T,0)>=0) printf("my teacher!\n"); else printf("senior!\n"); } } return 0; }
E. XKC's basketball (monotone queue)
The meaning of problems: the length of the array 5e5, for each of a [i], a right of the last request> = (a [i] + m) subscript.
Ideas: lx written. Reverse traversal, incrementing a queue maintained, not only the team into the team, if the current a [i] <= que [end], then not into the team, since than a [i] and after a large number of i will block a [i] effect; when a [i]> que [end], is added to the tail. Each query may traverse obtained que [begin] ~ que [end] the first> = a [i] + subscript m is, of course, also be used to accelerate half.
AC Code:
#include <iostream> #include <cstdio> #include <stack> #include <map> using namespace std; inline int read() { int x=0,f=1; char ch = 0; while(!isdigit(ch)) { if(ch=='-') f = -1; ch = getchar(); } while(isdigit(ch)) { x = x*10+ch-'0'; ch = getchar(); } return x*f; } int n,m; long long num[500050]; int ans[500050]; int dui[500050]; int ed; int main() { n = read(), m = read(); for(int i=1; i<=n; i++) num[i] = read(); for(int i=n; i>=1; i--) { int you = 0; while(you<ed && num[i]+m>num[dui[you]]) you++; if(you>=ed) { ans[i] = -1; if(num[i]>num[dui[ed-1]]) { dui[ed] = i; ed++; } } else { ans[i] = dui[you]-i-1; } } for(int i=1; i<=n; i++) { if(i>1) putchar(' '); printf("%d",ans[i]); } putchar('\n'); return 0; }
G. Colorful String (palindrome automata)
Meaning of the questions: seeking to set the value of all palindromes sub-string string, the string value is defined as the number of different characters.
Ideas: palindrome automaton naked. Competition site to learn palindrome automatic machine obtains the interval and the number of all kinds of palindromes palindromic sequence with an automatic machine. Then get the value of all the recording prefix substring palindromic calculation.
AC code：
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long LL; const int maxn=3e5+5; char s[maxn],s1[maxn]; int n,p,q,fail[maxn],cnt[maxn],len[maxn],tot,last,ch[maxn][26],l[maxn],r[maxn]; int pre[maxn][30],leng; LL ans; inline int newnode(int x){ len[++tot]=x; return tot; } inline int getfail(int x,int n){ while(s[n-len[x]-1]!=s[n]) x=fail[x]; return x; } int main(){ scanf("%s",s+1); leng=strlen(s+1); for(int i=1;i<=leng;++i) s1[i]=s[i]; s[0]=-1,fail[0]=1,last=0; len[0]=0,len[1]=-1,tot=1; for(int i=1;s[i];++i){ s[i]-='a'; p=getfail(last,i); if(!ch[p][s[i]]){ q=newnode(len[p]+2); l[q]=i-len[q]+1,r[q]=i; fail[q]=ch[getfail(fail[p],i)][s[i]]; ch[p][s[i]]=q; } ++cnt[last=ch[p][s[i]]]; } for(int i=tot;i;--i) cnt[fail[i]]+=cnt[i]; for(int i=1;i<=leng;++i){ for(int j=0;j<26;++j) pre[i][j]=pre[i-1][j]; ++pre[i][s1[i]-'a']; } for(int i=tot;i;--i){ int tmp=0; for(int j=0;j<26;++j) if(pre[r[i]][j]-pre[l[i]-1][j]>0) ++tmp; ans+=1LL*tmp*cnt[i]; } printf("%lld\n",ans); return 0; }
K. Center (computational geometry)
Meaning of the questions: n points to set point, find the minimum number of points added that all point to a point on the center of symmetry.
Thinking: n ^ 2 enumerate all the points of the midpoint, to find the most central point of enumeration number, which is the most appropriate point of the center point, assuming that the number of times it appears as Max, the point tmp, then the answer is (n- 2 * Max + mp1 [tmp]), mp1 used to record the number of points of the original focus point. For ease of handling, it may be the initial point × 2, so that the center point of a certain point is an integer.
AC eat oh:
#include<cstdio> #include<algorithm> #include<cmath> #include<cstdlib> #include<map> #include<utility> using namespace std; typedef pair<int,int> PII; const int maxn=1005; int n,Max; PII pt[maxn],tmp; map<PII,int> mp1,mp2; int main(){ scanf("%d",&n); for(int i=1;i<=n;++i){ scanf("%d%d",&pt[i].first,&pt[i].second); pt[i].first*=2,pt[i].second*=2; ++mp1[pt[i]]; } for(int i=1;i<=n;++i) for(int j=i;j<=n;++j){ PII t=make_pair((pt[i].first+pt[j].first)/2,(pt[i].second+pt[j].second)/2); ++mp2[t]; if(mp2[t]>Max){ Max=mp2[t]; tmp=t; } else if(mp2[t]==Max){ if(mp1[t]<mp1[tmp]) tmp=t; } } printf("%d\n",n-2*Max+mp1[tmp]); return 0; }