XVII Open Cup named after E.V. Pankratiev. Eastern GP, Division 1
A. Count The Ones
$ans=b-c+1$。
#include <stdio.h>
using namespace std ;
int a , b , c ;
void solve () {
printf ( "%d\n" , 1 + b - c ) ;
}
int main () {
while ( ~scanf ( "%d%d%d" , &a , &b , &c ) ) solve () ;
return 0 ;
}
B. Craters
求出凸包，然后枚举凸包上两个点，对第三个点旋转卡壳。因为随机数据凸包期望点数为$O(\sqrt{n})$，故时间复杂度为$O(n\log n)$。
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<int,int>PI;
const int N=200010;
int n,m,_,i,j,k,x,y;
set<PI>T;
ll ans=-1;
struct P{
int x,y;
P(){}
P(int _x,int _y){x=_x,y=_y;}
P operator-(P b){return P(x-b.x,y-b.y);}
bool operator<(const P&p)const{
if(x!=p.x)return x<p.x;
return y<p.y;
}
void write(){
printf("%d %d\n",x,y);
}
}a[N],b[N<<1],ans0,ans1,ans2;
inline ll cross(P a,P b){
return 1LL*a.x*b.y-1LL*a.y*b.x;
}
inline ll vect(P p,P p1,P p2){
return 1LL*(p1.x-p.x)*(p2.y-p.y)-1LL*(p1.y-p.y)*(p2.x-p.x);
}
int convexhull(P*p,int n,P*q){
int i,k,m;
sort(p,p+n);
m=0;
for(i=0;i<n;q[m++]=p[i++])while(m>1&&vect(q[m-2],q[m-1],p[i])<=0)m--;
k=m;
for(i=n-2;i>=0;q[m++]=p[i--])while(m>k&&vect(q[m-2],q[m-1],p[i])<=0)m--;
return --m;
}
inline void work(P a,P b,P c){
ll now=cross(a,b)+cross(b,c)+cross(c,a);
if(now<0)now=-now;
if(now>ans)ans=now,ans0=a,ans1=b,ans2=c;
}
inline ll area(P a,P b,P c){
ll now=cross(a,b)+cross(b,c)+cross(c,a);
if(now<0)now=-now;
return now;
}
int main(){
scanf("%d",&n);
if(n<200){
for(i=0;i<n;i++)scanf("%d%d",&a[i].x,&a[i].y);
for(i=0;i<n;i++)for(j=0;j<i;j++)for(k=0;k<j;k++){
work(a[i],a[j],a[k]);
}
ans0.write();
ans1.write();
ans2.write();
return 0;
}
_=n;
n=0;
while(_--){
scanf("%d%d",&x,&y);
if(T.find(PI(x,y))!=T.end())continue;
a[n++]=P(x,y);
T.insert(PI(x,y));
}
m=convexhull(a,n,b);
for(i=0;i<m;i++)b[i+m]=b[i];
for(i=0;i<m;i++){
for(j=i+1,k=j+1;j<i+m;j++){
if(k<=j)k=j+1;
while(k+1<i+m&&area(b[i],b[j],b[k])<=area(b[i],b[j],b[k+1]))k++;
work(b[i],b[j],b[k]);
}
}
ans0.write();
ans1.write();
ans2.write();
}
C. MSTrikes back!
记录最后$5$个点连通性的最小表示然后DP，注意到边权的周期不会太大，所以可以直接跳过完整的周期。
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<time.h>
#include<assert.h>
#include<iostream>
using namespace std;
typedef long long LL;
typedef pair<int,int>pi;
const int Ms=3125;
const LL Inf=1LL<<60;
const int LIM=100;
int n,seed;
LL in[LIM+1];
LL f[LIM+1][Ms];
LL dp[2][Ms];
void init(int cs,LL val){
for(int i=0;i<Ms;i++)dp[cs][i]=val;
}
int g[Ms][(1<<5)+1];
void decode(int mask,int *a){
for(int i=0;i<5;i++){
a[i]=mask%5;
mask/=5;
}
}
void zuixiao(int *a){
int dui[10];
memset(dui,-1,sizeof dui);
int cur=0;
for(int i=0;i<5;i++){
int x=a[i];
if(dui[x]<0){
dui[x]=cur++;
}
a[i]=dui[x];
}
}
int encode(int *a){
int res=0,tmp=1;
for(int i=0;i<5;i++,tmp=tmp*5){
res+=a[i]*tmp;
}
return res;
}
bool testt(int *a){
int b[10];
for(int i=0;i<5;i++)b[i]=a[i];
zuixiao(b);
for(int i=0;i<5;i++)if(b[i]!=a[i])return 0;
return 1;
}
void pre(){
int a[10],b[10];
for(int mask=0;mask<Ms;mask++){
decode(mask,a);
if(!testt(a)){
for(int j=0;j<1<<5;j++)g[mask][j]=-1;
continue;
}
//for(int i=0;i<5;i++)printf("%d ",a[i]);puts(":\n");
for(int j=0;j<1<<5;j++){
for(int i=0;i<5;i++)b[i]=a[i];
b[5]=6;
for(int k=0;k<5;k++){
if(j>>k&1){
for(int t=0;t<5;t++){
if(t==k)continue;
if(b[t]==b[k])b[t]=b[5];
}
b[k]=b[5];
}
}
bool flag=0;
for(int k=1;k<=5;k++)if(b[0]==b[k]){flag=1;break;}
if(!flag)g[mask][j]=-1;
else{
for(int k=0;k<5;k++)b[k]=b[k+1];
zuixiao(b);
//for(int k=0;k<5;k++)printf("%d ",b[k]);puts("");
g[mask][j]=encode(b);
}
}
}
}
void upd(LL &x,LL y){if(x>y)x=y;}
int main(){
pre();
int _;scanf("%d",&_);
while(_--){
scanf("%d%d",&n,&seed);
int cs=0;
init(cs,Inf);
dp[cs][0]=0;
for(int i=2;i<=n&&i<=LIM;i++){
init(cs^1,Inf);
seed=seed*907%2333333;
LL bq[10];
int T=seed;
in[i]=T;
for(int j=0;j<5;j++){
if(i-(5-j)<1)bq[j]=Inf;
else {
seed=seed*907%2333333;
bq[j]=seed^T;
}
}
for(int mask=0;mask<Ms;mask++){
LL w=dp[cs][mask];
if(w==Inf)continue;
for(int j=0;j<1<<5;j++){
if(g[mask][j]<0)continue;
bool flag=1;
for(int k=0;k<5;k++)if((j>>k&1)&&(bq[k]==Inf)){
flag=0;continue;
}
if(!flag)continue;
LL nw=w;
for(int k=0;k<5;k++){
if(j>>k&1)nw+=bq[k];
}
upd(dp[cs^1][g[mask][j]],nw);
}
}
cs^=1;
for(int j=0;j<Ms;j++)f[i][j]=dp[cs][j];
}
if(n<=LIM)printf("%lld\n",dp[cs][0]);
else{
LL pace,pacew;
for(int i=LIM-1;;i--)if(in[i]==in[LIM]){
pace=LIM-i;
break;
}
int st=(n-LIM)/pace*pace+LIM;
//printf("st=%d\n",st);
for(int i=0;i<Ms;i++){
dp[cs][i]=f[LIM][i]+1LL*(n-LIM)/pace*(f[LIM][i]-f[LIM-pace][i]);
}
for(int i=st+1;i<=n;i++){
init(cs^1,Inf);
seed=seed*907%2333333;
LL bq[10];
int T=seed;
for(int j=0;j<5;j++){
if(i-(5-j)<1)bq[j]=Inf;
else {
seed=seed*907%2333333;
bq[j]=seed^T;
}
}
for(int mask=0;mask<Ms;mask++){
LL w=dp[cs][mask];
if(w==Inf)continue;
for(int j=0;j<1<<5;j++){
if(g[mask][j]<0)continue;
bool flag=1;
for(int k=0;k<5;k++)if((j>>k&1)&&(bq[k]==Inf)){
flag=0;continue;
}
if(!flag)continue;
LL nw=w;
for(int k=0;k<5;k++){
if(j>>k&1)nw+=bq[k];
}
upd(dp[cs^1][g[mask][j]],nw);
}
}
cs^=1;
}
printf("%lld\n",dp[cs][0]);
}
}
return 0;
}
D. Skyscrapers
首先用set维护所有已经被摧毁的建筑。然后用线段树维护区间内最容易被向左/向右的冲击波击毁的建筑即可。时间复杂度$O(n\log n)$。
#include<cstdio>
#include<set>
#include<algorithm>
using namespace std;
const int N=100010,M=262150;
int n,m,i,x,ans,f[N],g[N];
set<int>T;
int vl[M],vr[M],pos[N],p;
inline int mergel(int x,int y){
if(!x||!y)return x+y;
return f[x]>f[y]?x:y;
}
inline int merger(int x,int y){
if(!x||!y)return x+y;
return g[x]<g[y]?x:y;
}
void build(int x,int a,int b){
if(a==b){
pos[a]=x;
vl[x]=vr[x]=a;
return;
}
int mid=(a+b)>>1;
build(x<<1,a,mid);
build(x<<1|1,mid+1,b);
vl[x]=mergel(vl[x<<1],vl[x<<1|1]);
vr[x]=merger(vr[x<<1],vr[x<<1|1]);
}
void askl(int x,int a,int b,int c,int d){
if(c<=a&&b<=d){
p=mergel(p,vl[x]);
return;
}
int mid=(a+b)>>1;
if(c<=mid)askl(x<<1,a,mid,c,d);
if(d>mid)askl(x<<1|1,mid+1,b,c,d);
}
void askr(int x,int a,int b,int c,int d){
if(c<=a&&b<=d){
p=merger(p,vr[x]);
return;
}
int mid=(a+b)>>1;
if(c<=mid)askr(x<<1,a,mid,c,d);
if(d>mid)askr(x<<1|1,mid+1,b,c,d);
}
inline void kill(int x){
ans++;
T.insert(x);
x=pos[x];
vl[x]=vr[x]=0;
for(x>>=1;x;x>>=1){
vl[x]=mergel(vl[x<<1],vl[x<<1|1]);
vr[x]=merger(vr[x<<1],vr[x<<1|1]);
}
}
inline void workl(int l,int r,int t){
if(l>r)return;
while(1){
p=0;
askl(1,1,n,l,r);
if(!p)return;
if(f[p]<t)return;
kill(p);
}
}
inline void workr(int l,int r,int t){
if(l>r)return;
while(1){
p=0;
askr(1,1,n,l,r);
if(!p)return;
if(g[p]>t)return;
kill(p);
}
}
int main(){
scanf("%d",&n);
for(i=1;i<=n;i++)scanf("%d",&x),f[i]=i-x,g[i]=i+x;
T.insert(0);
T.insert(n+1);
build(1,1,n);
scanf("%d",&m);
while(m--){
scanf("%d",&x);
ans=0;
kill(x);
set<int>::iterator j=T.find(x),k;
k=j;
j--;k++;
workl(*j+1,x-1,f[x]);
workr(x+1,*k-1,g[x]);
printf("%d\n",ans);
}
/*
left:
a[x]-a[y]>=x-y
y-a[y]>=x-a[x]
get max(y-a[y])
right:
a[x]-a[y]>=y-x
y+a[y]<=x+a[x]
get min(y+a[y])
*/
}
E. Blackboard
按题意模拟即可。
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<time.h>
#include<assert.h>
#include<iostream>
using namespace std;
typedef long long LL;
typedef pair<int,int>pi;
int n,ty;
int a[103][103];
void rev(){
for(int i=1;i<=n;i++){
for(int j=i;j<=n;j++){
swap(a[i][j],a[j][i]);
}
}
}
void solve1(){
int cur=1;
for(int i=1;i<=n;i++){
if(i&1){
for(int j=1;j<=n;j++){
a[i][j]=cur++;
}
}
else{
for(int j=n;j>=1;j--){
a[i][j]=cur++;
}
}
}
}
bool isok(int x,int y){
return x>=1&&x<=n&&y>=1&&y<=n&&a[x][y]==0;
}
int di[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
void solve2(){
int d=0;
int sx=1,sy=1;
memset(a,0,sizeof a);
int cur=1;
for(int it=1;it<=n*n;it++){
a[sx][sy]=cur++;
int nx=sx+di[d][0],ny=sy+di[d][1];
if(it==n*n)break;
while(!isok(nx,ny)){
d++;
d%=4;
nx=sx+di[d][0],ny=sy+di[d][1];
}
sx=nx,sy=ny;
}
}
int main(){
while(scanf("%d%d",&n,&ty)!=EOF){
if(ty==1||ty==2)solve1();
if(ty==3||ty==4)solve2();
if(ty==2||ty==4)rev();
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
printf("%d%c",a[i][j],j==n?'\n':' ');
}
}
}
return 0;
}
F. Buddy Numbers
若质数太多那么显然无解，对于小数据爆搜即可。
#include <bits/stdc++.h>
using namespace std ;
int a[1000] , n ;
void solve () {
if ( n == 1 ) {
printf ( "1\n" ) ;
return ;
}
if ( n == 2 ) {
printf ( "1 2\n" ) ;
return ;
}
if ( n == 3 ) {
printf ( "2 1 3\n" ) ;
return ;
}
if ( n == 4 ) {
printf ( "2 4 1 3\n" ) ;
return ;
}
if ( n == 6 ) {
printf ( "3 6 2 4 1 5\n" ) ;
return ;
}
printf ( "-1\n" ) ;
}
int main () {
while ( ~scanf ( "%d" , &n ) ) solve () ;
return 0 ;
}
G. Gmoogle
按题意模拟即可。注意首尾空格的处理以及文末没有标点符号的情况的处理。
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<time.h>
#include<assert.h>
#include<sstream>
#include<iostream>
using namespace std;
typedef long long LL;
typedef pair<int,int>pi;
string s;
vector<string>jz;
int can[1020];
int isend(int loc){
if(s[loc]!='.'&&s[loc]!='!'&&s[loc]!='?')return 0;
if(s[loc]=='.'){
int j=loc+1;
while(j<s.size()&&s[j]==' ')j++;
if(j<s.size()&&s[j]>='a'&&s[j]<='z')return 0;
}
return 1;
}
void cg(char &c){if(c>='A'&&c<='Z')c=c-'A'+'a';}
bool cmp(char c1,char c2){
cg(c1),cg(c2);
if(c1==c2)return 1;
return 0;
}
bool ok(string &wb,string& in){
for(int i=0;i+in.size()<=wb.size();i++)if(isalpha(wb[i])){
bool flag=1;
if(i&&isalpha(wb[i-1]))continue;//
int j=i+in.size();
if(j<wb.size()&&isalpha(wb[j]))continue;//
//while ( j < wb.size () && wb[j] == ' ' ) ++ j ;
//if(j<wb.size()-1&&wb[j]=='.')continue;
for(int j=0;j<in.size();j++){
if(cmp(wb[i+j],in[j])!=1){flag=0;break;}
}
if(flag)return 1;
}
return 0;
}
bool zhu(char x){
if(x>='a'&&x<='z')return 1;
if(x>='A'&&x<='Z')return 1;
if(x==' ')return 1;
if(x>='0'&&x<='9')return 1;
if(x=='.'||x=='?'||x=='!')return 1;
return 0;
}
void fix ( string& s ) {
while ( !zhu(s[s.size () - 1]) ) s.pop_back () ;
}
int main(){
getline(cin,s);
fix ( s ) ;
int k = 0 ;
for(int l=0;l<s.size();){
while(l<s.size()&&s[l]==' ')l++;
if(l>=s.size())break;
//if ( s[l] == '.' /*|| s[l] == '?' || s[k] == '!'*/ ) while ( 1 ) ;
int r=l;
string tmp="";
while(r<s.size()&&!isend(r))tmp.push_back(s[r++]);
if ( r < s.size () ) tmp.push_back(s[r++]);
while(tmp[tmp.size()-1]==' ')tmp.pop_back();
//printf ( "%d\n" , ( int ) tmp.size () ) ;
jz.push_back(tmp);
l=r;
}
/*
cout<<"OUTPUT"<<endl;
for(int i=0;i<jz.size();i++)cout<<jz[i]<<endl;
*/
int q;scanf("%d",&q);
string rub;
getline(cin,rub);
while(q--){
string que;
getline(cin,que);
fix ( que ) ;
for(int i=0;i<jz.size();i++)can[i]=1;
stringstream ss(que);
string tmp;
while(ss>>tmp){
fix ( tmp ) ;
for(int i=0;i<jz.size();i++){
if(!ok(jz[i],tmp))can[i]=0;
}
}
int i=0,j=que.size()-1;
while(que[i]==' ')i++;
while(que[j]==' ')j--;
cout<<"Search results for \"";
for(int it=i;it<=j;it++)printf("%c",que[it]);
cout<<"\":\n";
for(int i=0;i<jz.size();i++)if(can[i]){
cout<<"- \""<<jz[i]<<"\"\n";
}
}
return 0;
}
H. Generator
线性筛求出每个数的最小质因子即可。
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<time.h>
#include<assert.h>
#include<iostream>
using namespace std;
typedef long long LL;
typedef pair<int,int>pi;
LL n;
LL sum[10000020];
int tot[10000020];
bool isp[10000200];
vector<int>pri;
const int Maxn=10000010;
int getsq(int x){
int tmp=sqrt(x+0.5);
while(tmp*tmp<x)tmp++;
while(tmp*tmp>x)tmp--;
return tmp-1;
}
void pre(){
for(int i=2;i<Maxn;i++){
if(!isp[i])pri.push_back(i),sum[i]=getsq(i);
for(int j=0;j<pri.size();j++){
int x=pri[j];
if(x*i>=Maxn)break;
sum[x*i]=x-1;
isp[i*x]=1;
if(i%x==0)break;
}
}
//for(int i=2;i<=10;i++)printf("%lld ",sum[i]);puts("");
for(int i=2;i<Maxn;i++)sum[i]+=sum[i-1];
for(int i=2;i<Maxn;i++)tot[i]=tot[i-1]+(!isp[i]);
}
int main(){
pre();
int _;scanf("%d",&_);
while(_--){
scanf("%lld",&n);
LL ans=sum[n];
if(ans==0)printf("0/1\n");
else{
LL gc=__gcd((LL)tot[n],ans);
printf("%lld/%lld\n",ans/gc,tot[n]/gc);
}
}
return 0;
}
I. Addition
维护二进制字典树即可。
#include <bits/stdc++.h>
using namespace std ;
typedef long long LL ;
const int MAXN = 100005 ;
int nxt[MAXN][2] ;
int cnt[MAXN] ;
char s[MAXN] , s1[MAXN] ;
int cur , root ;
int n , m ;
int newnode () {
++ cur ;
nxt[cur][0] = nxt[cur][1] = 0 ;
cnt[cur] = 0 ;
return cur ;
}
void insert ( char s[] ) {
int now = root ;
for ( int i = 0 ; i < m ; ++ i ) {
int x = s[i] ;
if ( !nxt[now][x] ) nxt[now][x] = newnode () ;
now = nxt[now][x] ;
cnt[now] ++ ;
}
}
int query ( char s[] ) {
int now = root , ans = 0 ;
for ( int i = 0 ; i < m ; ++ i ) {
int x = s[i] ;
if ( !x ) ans += cnt[nxt[now][1]] ;
now = nxt[now][x] ;
}
return ans ;
}
void solve () {
cur = 0 ;
root = newnode () ;
for ( int i = 1 ; i <= n ; ++ i ) {
scanf ( "%s" , s ) ;
for ( int j = 0 ; j < m ; ++ j ) {
s[j] -= '0' ;
s1[j] = 1 ^ s[j] ;
}
int ans = query ( s1 ) ;
insert ( s ) ;
printf ( "%d\n" , ans ) ;
fflush ( stdout ) ;
}
}
int main () {
scanf ( "%d%d" , &n , &m ) ;
solve () ;
return 0 ;
}
J. Votter and Paul De Mort
留坑。
K. GCD on the segments
考虑枚举左端点$i$，则随着右端点的右移，一共只有$O(\log n)$种不同的$\gcd$取值。所以首先通过ST表+二分查找预处理出$O(n\log n)$个四元组$(x,i,l,r)$，表示左端点为$i$，右端点取值范围在$[l,r]$内，且这一段的$\gcd$都为$x$。
将四元组按照$x$为第一关键字，$i$为第二关键字排序，对于相同的$x$一起处理。
当$x$相同时，显然所有的$i$互不相同。设$f[i]$为恰好以位置$i$为结尾的最优解，则对于一个四元组$(x,i,l,r)$，能更新它的最优解为区间$[1,i-1]$的最优值$+1$，然后用它更新区间$[l,r]$的$f[]$。用支持打标记的线段树维护即可。时间复杂度$O(n\log^2n)$。
#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int N=100010,K=17,P=1000000007,M=262145;
int T,n,m,i,j,x,y,l,r,mid,Log[N],val,f[K][N];
struct PI{
int x,i,l,r;
PI(){}
PI(int _x,int _i,int _l,int _r){x=_x,i=_i,l=_l,r=_r;}
}a[3000000];
inline bool cmp(PI a,PI b){return a.x==b.x?a.i<b.i:a.x<b.x;}
struct Num{
int x,y;
Num(){x=y=0;}
Num(int _x,int _y){x=_x,y=_y;}
inline Num operator+(Num b){
if(x<b.x)return b;
if(x>b.x)return Num(x,y);
return Num(x,(y+b.y)%P);
}
inline Num operator+(int _x){return Num(x+_x,y);}
inline Num operator-(int b){return Num(x,(long long)y*b%P);}
inline void operator+=(Num b){*this=*this+b;}
}tmp,v[M],tag[M],ans;
int pos[M];
inline void read(int&a){char c;while(!(((c=getchar())>='0')&&(c<='9')));a=c-'0';while(((c=getchar())>='0')&&(c<='9'))(a*=10LL)+=c-'0';}
inline int askgcd(int y){int k=Log[y-i+1];return __gcd(f[k][i],f[k][y-(1<<k)+1]);}
inline void clean(int x){
if(pos[x]<T)pos[x]=T,v[x]=tag[x]=Num();
}
inline void tag1(int x,Num y){
clean(x);
v[x]+=y;
tag[x]+=y;
}
inline void pb(int x){
if(tag[x].x){
tag1(x<<1,tag[x]);
tag1(x<<1|1,tag[x]);
tag[x]=Num();
}
}
inline void up(int x){
clean(x<<1),clean(x<<1|1);
v[x]=v[x<<1]+v[x<<1|1];
}
void change(int x,int a,int b,int c,int d){
clean(x);
if(c<=a&&b<=d){tag1(x,tmp);return;}
pb(x);
int mid=(a+b)>>1;
if(c<=mid)change(x<<1,a,mid,c,d);
if(d>mid)change(x<<1|1,mid+1,b,c,d);
up(x);
}
void ask(int x,int a,int b,int c,int d){
clean(x);
if(c<=a&&b<=d){tmp+=v[x];return;}
pb(x);
int mid=(a+b)>>1;
if(c<=mid)ask(x<<1,a,mid,c,d);
if(d>mid)ask(x<<1|1,mid+1,b,c,d);
up(x);
}
int main(){
for(i=2;i<=100000;i++)Log[i]=Log[i>>1]+1;
while(~scanf("%d",&n)){
m=0;
ans=Num();
for(i=1;i<=n;i++)read(f[0][i]);
int flag=0;
for(i=2;i<=n;i++)if(f[0][i]!=f[0][i-1]){flag=1;break;}
if(!flag){printf("%d 1\n",n);continue;}
for(j=1;j<K;j++)for(i=1;i+(1<<j-1)<=n;i++)f[j][i]=__gcd(f[j-1][i],f[j-1][i+(1<<j-1)]);
for(i=1;i<=n;i++)for(x=i;x<=n;x=y+1){
val=askgcd(y=x),l=x+1,r=n;
while(l<=r)if(askgcd(mid=(l+r)>>1)==val)l=(y=mid)+1;else r=mid-1;
a[++m]=PI(val,i,x,y);
}
sort(a+1,a+m+1,cmp);
for(i=1;i<=m;i++){
if(i==1||a[i].x!=a[i-1].x)T++;
tmp=Num();
if(a[i].i>1)ask(1,1,n,1,a[i].i-1);
if(!tmp.x)tmp=Num(1,1);else tmp.x++;
ans+=tmp-(a[i].r-a[i].l+1);
change(1,1,n,a[i].l,a[i].r);
}
printf("%d %d\n",ans.x,ans.y);
}
return 0;
}
L. Fibonacci Equation
分类讨论。
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<time.h>
#include<assert.h>
#include<iostream>
using namespace std;
typedef long long LL;
typedef pair<int,int>pi;
LL a[100];
LL x,y,z;
int main(){
/*
a[0]=0,a[1]=1;
for(int i=2;i<=30;i++)a[i]=a[i-1]+a[i-2];
for(int i=1;i<=25;i++){
printf("%lld %lld\n",a[i],a[i]*a[i]-4*a[i-1]*a[i+1]);
}
*/
while(scanf("%lld%lld%lld",&x,&y,&z)!=EOF){
if(x==0){printf("1\n");continue;}
if(y>x&&y>z){
if(y==3)printf("1\n");
else printf("2\n");
}
if(y<x&&y<z)printf("0\n");
if(y>min(x,z)&&y<max(x,z)){
if(y==1)printf("2\n");
else printf("0\n");
}
//printf("real=%lld\n",a[y]*a[y]-4*a[x]*a[z]);
}
return 0;
}
