codeforces #pi
codeforces #pi
A,B题:水题,略。
C题:求一个序列中形成以k为公比项数为3的等比数列的种类数。考虑中间项b,统计b/k在前面出现的次数pre[b/k]和b*k在后面出现的次数suf[b*k],相乘。统计pre可以从前往后扫,统计suf则用all减去pre即可。
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<set> #include<map> #include<string> #include<math.h> #include<cctype> #define ll long long #define REP(i,a,b) for(int (i)=(a);(i)<=(b);(i)++) #define REPP(i,a,b,t) for(int (i)=(a);(i)<=(b);(i)+=(t)) #define rep(i,a,b) for(int (i)=(a);(i)>=(b);(i)--) #define repp(i,a,b,t) for(int (i)=(a);(i)>=(b);(i)-=(t)) #define PII pair<int,int> #define fst first #define snd second #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&(x)) #define RII(x,y) scanf("%d%d",&(x),&(y)) #define RIII(x,y,z) scanf("%d%d%d",&(x),&(y),&(z)) #define DRI(x) int (x);scanf("%d",&(x)) #define DRII(x,y) int (x),(y);scanf("%d%d",&(x),&(y)) #define DRIII(x,y,z) int (x),(y),(z);scanf("%d%d%d",&(x),&(y),&(z)) #define RS(x) scanf("%s",x) #define RSS(x,y) scanf("%s%s",x,y) #define DRS(x) char x[maxn];scanf("%s",x) #define DRSS(x,y) char x[maxn],y[maxn];scanf("%s%s",x,y) #define MS0(a) memset((a),0,sizeof((a))) #define MS1(a) memset((a),-1,sizeof((a))) #define MS(a,b) memset((a),(b),sizeof((a))) #define ALL(v) v.begin(),v.end() #define SZ(v) (int)(v).size() using namespace std; const int maxn=1000100; const int INF=(1<<29); const double EPS=0.0000000001; const double Pi=acos(-1.0); int n,k; ll a[maxn]; map<ll,int> all,pre; int main() { //freopen("in.txt","r",stdin); while(~RII(n,k)){ all.clear();pre.clear(); REP(i,1,n){ scanf("%I64d",&a[i]); all[a[i]]++; } ll ans=0; REP(i,1,n){ if(a[i]%k==0){ ll l=pre[a[i]/k]; ll r=all[a[i]*k]-pre[a[i]*k]; if(k==1||a[i]==0) r--; ans+=l*r; } pre[a[i]]++; } cout<<ans<<endl; } return 0; }
D题:
注意set的二分,s.lower_bound(x),求比x小的数用s.lower_bound(x)-1即可。
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<set> #include<map> #include<string> #include<math.h> #include<cctype> #define ll long long #define REP(i,a,b) for(int (i)=(a);(i)<=(b);(i)++) #define REPP(i,a,b,t) for(int (i)=(a);(i)<=(b);(i)+=(t)) #define rep(i,a,b) for(int (i)=(a);(i)>=(b);(i)--) #define repp(i,a,b,t) for(int (i)=(a);(i)>=(b);(i)-=(t)) #define PII pair<int,int> #define fst first #define snd second #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&(x)) #define RII(x,y) scanf("%d%d",&(x),&(y)) #define RIII(x,y,z) scanf("%d%d%d",&(x),&(y),&(z)) #define DRI(x) int (x);scanf("%d",&(x)) #define DRII(x,y) int (x),(y);scanf("%d%d",&(x),&(y)) #define DRIII(x,y,z) int (x),(y),(z);scanf("%d%d%d",&(x),&(y),&(z)) #define RS(x) scanf("%s",x) #define RSS(x,y) scanf("%s%s",x,y) #define DRS(x) char x[maxn];scanf("%s",x) #define DRSS(x,y) char x[maxn],y[maxn];scanf("%s%s",x,y) #define MS0(a) memset((a),0,sizeof((a))) #define MS1(a) memset((a),-1,sizeof((a))) #define MS(a,b) memset((a),(b),sizeof((a))) #define ALL(v) v.begin(),v.end() #define SZ(v) (int)(v).size() using namespace std; const int maxn=1000100; const int INF=(1<<29); const double EPS=0.0000000001; const double Pi=acos(-1.0); int n,k,a; int m; int b; set<int> s; int x,y; int main() { //freopen("in.txt","r",stdin); while(~RIII(n,k,a)){ RI(m); s.clear(); int sum=(n+1)/(a+1); s.insert(0);s.insert(n+1); int ans=-1; bool flag=0; REP(i,1,m){ RI(b); set<int>::iterator it=s.upper_bound(b); int y=*it; int x=*(--it); sum-=(y-x)/(a+1)-(y-b)/(a+1)-(b-x)/(a+1); if(sum<k&&!flag) ans=i,flag=1; s.insert(b); } cout<<ans<<endl; } return 0; }
没有AC不了的题,只有不努力的ACMER!