1001（打表/不定方程）
这个可以直接打表。。然后发现。。如果不能被3或者4整除就是-1。。
然后被3整除一定是x=y=z=n/3（废话），否则被4整除一定是x=y=z/2=n/4。。
然后是严格证明。。
令u=n/x,v=n/y,w=n/z
有1/u+1/v+1/w=1
假设u<=v<=w,则u<=3
当u=2，1/v+1/w=1/2，v=w=4或w=3,v=6
当u=3，1/v+1/w=2/3，v=w=3
就只有这3种取法。。然后u=2,w=3,v=6如果能满足一定能满足3|n了。。直接取均值，可排除。。
/**
* ┏┓ ┏┓
┏┛┗━━━━━━━┛┗━━━┓
┃ ┃
┃ ━ ┃
┃ ＞ ＜ ┃
┃ ┃
┃... ⌒ ... ┃
┃ ┃
┗━┓ ┏━┛
┃ ┃ Code is far away from bug with the animal protecting
┃ ┃ 神兽保佑,代码无bug
┃ ┃
┃ ┃
┃ ┃
┃ ┃
┃ ┗━━━┓
┃ ┣┓
┃ ┏┛
┗┓┓┏━━━━━━━━┳┓┏┛
┃┫┫ ┃┫┫
┗┻┛ ┗┻┛
*/
include<cstdio>
include<cstring>
include<algorithm>
include<iostream>
include<queue>
include<cmath>
include<map>
include<stack>
include<set>
define inc(i,l,r) for(int i=l;i<=r;i++)
define dec(i,l,r) for(int i=l;i>=r;i--)
define link(x) for(edge *j=h[x];j;j=j->next)
define mem(a) memset(a,0,sizeof(a))
define ll long long
define eps 1e-12
define succ(x) (1<<x)
define lowbit(x) (x&(-x))
define sqr(x) ((x)*(x))
define mid (x+y>>1)
define NM 1000005
define nm 2000005
define N 1000005
define M(x,y) x=max(x,y)
const double pi=acos(-1);
const ll inf=998244353;
using namespace std;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x10+ch-'0',ch=getchar();
return fx;
}
ll n;
int main(){
int =read();while(--){
n=read();
if(n%3==0)n/=3,printf("%lld
",nnn);
else if(n%4==0)n/=4,printf("%lld
",nnn*2);
else printf("-1
");
}
return 0;
}
1002(6299)Balanced Sequence
题解:可以通过类似栈的操作把序列化成')))))','((((((',')))))((((('这样的三种形式，然后对于这个pair(a,b)按照玄学规则排序就OK(G++T了C++就过了)
#include <iostream>
#include <algorithm>
#include <string>
const int MAXN=1e5+10;
const int maxn=5e6+10;
using namespace std;
string s[MAXN];
char str[maxn];
typedef struct node{
int h1,h2,id;
friend bool operator<(node aa,node bb){
if(min(aa.h1,bb.h2)==min(aa.h2,bb.h1))return aa.h1<bb.h2;
return min(aa.h1,bb.h2)>min(aa.h2,bb.h1);
}
}node;
node d[MAXN];
/*bool cmp(node aa,node bb){
if(min(aa.h1,bb.h2)==min(aa.h2,bb.h1))return aa.h1<bb.h2;
return min(aa.h1,bb.h2)>min(aa.h2,bb.h1);
}*/
int main(){
//int _;cin>>_;
ios::sync_with_stdio(false);
int _;cin>>_;
while(_--){
int n;cin>>n;
for(int i=1;i<=n;i++)cin>>s[i];
int ans=0;
for(int i=1;i<=n;i++){
int _t=s[i].size();
d[i].h1=d[i].h2=0;d[i].id=i;
for(int j=0;j<_t;j++){
if(s[i][j]=='(')d[i].h1++;
else{
if(d[i].h1>0)d[i].h1--;
else{d[i].h2++;}
}
}
}
sort(d+1,d+n+1);
// sort(d+1,d+n+1);
int cnt=0;
for(int i=1;i<=n;i++){
int _t=s[d[i].id].length();
for(int j=0;j<_t;j++)str[++cnt]=s[d[i].id][j];
}
int a1=0,b1=0;
for(int i=1;i<=cnt;i++){
if(str[i]=='(')a1++;
else{
if(a1)a1--,ans++;
else b1++;
}
/* ans+=2*min(b1,d[i].h2);
b1-=min(b1,d[i].h2);
b1+=d[i].h1;*/
}
ans*=2;
cout<<ans<<endl;
// for(int i=1;i<=n;i++)s[i].clear();
}
return 0;
}
1003(模拟）
模拟一下发现可以从左往右取，如果有x相同的可以取y最大或者y最小的。
/**
* ┏┓ ┏┓
┏┛┗━━━━━━━┛┗━━━┓
┃ ┃
┃ ━ ┃
┃ ＞ ＜ ┃
┃ ┃
┃... ⌒ ... ┃
┃ ┃
┗━┓ ┏━┛
┃ ┃ Code is far away from bug with the animal protecting
┃ ┃ 神兽保佑,代码无bug
┃ ┃
┃ ┃
┃ ┃
┃ ┃
┃ ┗━━━┓
┃ ┣┓
┃ ┏┛
┗┓┓┏━━━━━━━━┳┓┏┛
┃┫┫ ┃┫┫
┗┻┛ ┗┻┛
*/
include<cstdio>
include<cstring>
include<algorithm>
include<iostream>
include<queue>
include<cmath>
include<map>
include<stack>
include<set>
define inc(i,l,r) for(int i=l;i<=r;i++)
define dec(i,l,r) for(int i=l;i>=r;i--)
define link(x) for(edge *j=h[x];j;j=j->next)
define mem(a) memset(a,0,sizeof(a))
define ll long long
define eps 1e-12
define succ(x) (1<<x)
define lowbit(x) (x&(-x))
define sqr(x) ((x)*(x))
define mid (x+y>>1)
define NM 1000005
define nm 2000005
define N 1000005
define M(x,y) x=max(x,y)
const double pi=acos(-1);
const ll inf=998244353;
using namespace std;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x10+ch-'0',ch=getchar();
return fx;
}
struct P{
int x,y,i;
bool operator<(const P&o){return x<o.x||(x==o.x&&y<o.y);}
}a[NM];
int n;
int main(){
int =read();while(--){
n=read()*3;
inc(i,1,n)a[i].x=read(),a[i].y=read(),a[i].i=i;
sort(a+1,a+1+n);
inc(i,1,n){
if(i%3)printf("%d ",a[i].i);else printf("%d
",a[i].i);
}
}
}
1004(6301)Distinct Values
题解:找到每个位置被限制的最左端点的位置 然后求这个区间的mex 弱鸡写了主席树做法 双制作+set的并没有想到
#include <bits/stdc++.h>
const int MAXN=1e5+10;
using namespace std;
typedef struct node{
int l,r,vis;int len;
}node;
node d[MAXN*21];int rt[MAXN];
int cnt;
void update(int &x,int y,int l,int r,int t){
x=++cnt;d[x]=d[y];d[x].len++;
// cout<<l<<"===----"<<r<<endl;
if(l==r){
//if(!d[x].vis)d[x].len=1,d[x].vis=1;
return ;
}
int mid=(l+r)>>1;
if(t<=mid)update(d[x].l,d[y].l,l,mid,t);
else update(d[x].r,d[y].r,mid+1,r,t);
d[x].len=d[d[x].l].len+d[d[x].r].len;
// cout<<l<<" "<<r<<" "<<d[x].len<<endl;
}
int ans;
void querty(int x,int y,int l,int r){
if(l==r){ans=l;return ;}
int mid=(l+r)>>1;
// cout<<l<<"===="<<r<<" "<<d[d[y].l].len<<" "<<d[d[x].l].len<<endl;
if(d[d[y].l].len-d[d[x].l].len!=mid-l+1)querty(d[x].l,d[y].l,l,mid);
else querty(d[x].r,d[y].r,mid+1,r);
}
typedef struct Q{
int l,r;
friend bool operator<(Q aa,Q bb){
if(aa.l==bb.l)return aa.r>bb.r;
return aa.l<bb.l;
}
}Q;
Q que[MAXN];
int ans1[MAXN],pre[MAXN];
int main(){
int _;scanf("%d",&_);
while(_--){
int n,m;scanf("%d%d",&n,&m);
cnt=0;
for(int i=1;i<=m;i++)scanf("%d%d",&que[i].l,&que[i].r);
sort(que+1,que+m+1);
int be=1;
for(int i=1;i<=m;i++){
while(be<que[i].l)pre[be]=0,ans1[be]=1,be++;
while(be<=que[i].r)pre[be]=que[i].l,be++;
}
for(;be<=n;be++)pre[be]=0,ans1[be]=1;
// for(int i=1;i<=n;i++)cout<<pre[i]<<" ";
// cout<<endl;
int l=1;ans=1;int r=0;
for(int i=1;i<=m;i++){
while(l<que[i].l)ans1[l]=1,rt[l]=rt[l-1],l++;
r=0;
while(l<=que[i].r){
querty(rt[que[i].l-1],rt[l-1],1,n);ans1[l]=ans;
// cout<<l<<":::::"<<ans<<endl;
update(rt[l],rt[l-1],1,n,ans);
// cout<<"--------"<<endl;
l++;
}
}
for(;l<=n;l++)ans1[l]=1;
for(int i=1;i<=n;i++){if(i==1)printf("%d",ans1[i]);else printf(" %d",ans1[i]);}
printf("
");
}
}
1007队友博客https://blog.csdn.net/qkoqhh/article/details/81181298
1008(6305) RMQ Similar Sequence
题解:http://tokitsukaze.live/2018/07/23/hdu6305/
大佬的博客有详细题解 简单说只要证明B数组是一个排列 后面的做法就很显然...分治+st表
/*#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#pragma comment(linker, "/STACK:1024000000,1024000000")*/
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int MAXN=1e6+10;
const ll mod=1e9+7;
//int n;
/*ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}*/
struct FastIO
{
static const int S=200;
int wpos;
char wbuf[S];
FastIO():wpos(0){}
inline int xchar()
{
static char buf[S];
static int len=0,pos=0;
if(pos==len) pos=0,len=fread(buf,1,S,stdin);
if(pos==len) exit(0);
return buf[pos++];
}
inline int read()
{
int s=1,c=xchar(),x=0;
while(c<=32) c=xchar();
if(c=='-') s=-1,c=xchar();
for(;'0'<=c&&c<='9';c=xchar()) x=x*10+c-'0';
return x*s;
}
~FastIO()
{
if(wpos) fwrite(wbuf,1,wpos,stdout),wpos=0;
}
}io;
ll ksm(ll a,ll b){
ll ans=1ll;
while(b){
if(b&1)ans=ans*a%mod;
a=a*a%mod;b>>=1;
}
return ans;
}
int a[MAXN];int pos[MAXN][22];
int mu[22],ma[MAXN];ll fav[MAXN],sum[MAXN];
int pmin(int aa,int bb){
if(a[aa]==a[bb])return min(aa,bb);
return a[aa]>a[bb]?aa:bb;
}
void St(int n){
for(int i=1;i<=n;i++){pos[i][0]=i;
for(int j=1;i+(1<<(j-1))<=n+1;j++)pos[i][j]=0;
}
for(int j=1;mu[j-1]<=n;j++){
for(int i=1;i+mu[j]<=n+1;i++){
int t=mu[j-1];
// if(dp[i][j-1]==dp[i+mu[j-1]][j-1])dp[i][j]=dp[i][j-1],pos[i][j]=min(dp[i][j-1],dp[i+mu[j-1]][j-1]);
// else if(dp[i][j-1]>dp[i+mu[j-1]][j-1])dp[i][j]=dp[i][j-1],pos[i][j]=pos[i][j-1];
// else dp[i][j]=dp[i+mu[j-1]][j-1],pos[i][j]=pos[i+mu[j-1]][j-1];
pos[i][j]=pmin(pos[i][j-1],pos[i+t][j-1]);
}
}
}
int rmq(int l,int r){
// int k=ma[r-l+1];
int j=(int)(log10(r-l+1)/log10(2))+1;
int i=r-(1<<(j-1))+1;
return pmin(pos[l][j-1],pos[i][j-1]);
// if(dp[l][k]==dp[r-mu[k]+1][k])return min(pos[l][k],pos[r-mu[k]+1][k]);
// else if(dp[l][k]>dp[r-mu[k]+1][k])return pos[l][k];
// else return pos[r-mu[k]+1][k];
// return pmin(pos[l][k],pos[r-mu[k]+1][k]);
}
ll C(int nn,int mm){
//ll ret=1;
if(mm<nn||nn<0) return 0;
return fav[nn]*fav[mm-nn]%mod*sum[mm]%mod;
}
namespace DFS
{
struct node{int l,r;};
int top;
node st[MAXN];
ll dfs(int mn,int mx)
{
ll res=1;
int l,r;
l=mn;
r=mx;
top=0;
st[top].l=l;
st[top++].r=r;
while(top)
{
l=st[top-1].l;
r=st[--top].r;
if(l>=r) continue;
int pos=rmq(l,r);
// res*=fav[p-l];res%=mod;res*=fav[r-p];res%=mod;res*=sum[r-l];res%=mod;
res=res*C(pos-l,r-l)%mod;
st[top].l=l;
st[top++].r=pos-1;
st[top].l=pos+1;
st[top++].r=r;
}
return res;
}
}
/*ll C(ll n,ll m){
if(n>m) return 0;
ll a=1,b=1;
for(int i=m;i>=m-n+1;i--) a=a*i%mod;
for(int i=1;i<=n;i++) b=b*i%mod;
return (a*(ksm(b,mod-2)%mod)%mod);
}
ll Lucas(ll n,ll m){
if(n==0) return 1;
return (1ll*C(n%mod,m%mod)*Lucas(n/mod,m/mod))%mod;
}*/
int p;
ll dfs(int l,int r){
//cout<<l<<" "<<r<<endl;
if(l>=r)return 1LL;
// cout<<l<<" "<<r<<" "<<p<<endl;
p=rmq(l,r);
// cout<<l<<" "<<r<<" "<<p<<" "<<fav[p-l]<<" "<<res<<endl;
// cout<<res<<endl;
return dfs(l,p-1)*dfs(p+1,r)%mod*C(p-l,r-l)%mod;
//dfs(rmq(l,r)+1,r);
}
int main(){
int _;int n;
_=io.read();
mu[0]=1;//ma[0]=-1;
// scanf("%d",&_);
//cout<<_<<endl;
for(int i=1;i<=21;i++)mu[i]=mu[i-1]*2;
// for(int i=1;i<MAXN;i++)if(!(i&(i-1)))ma[i]=ma[i-1]+1;else ma[i]=ma[i-1];
sum[0]=1;fav[0]=1;
for(int i=1;i<MAXN;i++)sum[i]=i*sum[i-1]%mod,fav[i]=ksm(sum[i],mod-2);
//cout<<Lucas(2,3)<<endl;
while(_--){
n=io.read();
// scanf("%d",&n);
// memset(dp,0,sizeof(dp));memset(pos,0,sizeof(dp));
for(int i=1;i<=n;i++)a[i]=io.read();
St(n);ll ans=DFS::dfs(1,n);
// cout<<res<<endl;
ans*=ksm((sum[n])%mod,mod-2);ans%=mod;
ans*=n;ans%=mod;ans*=ksm(2ll,mod-2);ans%=mod;
printf("%lld
",ans);
}
return 0;
}
1011(6308)Time Zone
题解:模拟
#include <bits/stdc++.h>
using namespace std;
int main(){
int n;scanf("%d",&n);double t;int x,y;
while(n--){
char ch='1';
scanf("%d%d",&x,&y);
while((ch!='+'&&ch!='-'))ch=getchar();
scanf("%lf",&t);
if(ch=='-')t*=-1;
t-=8;bool flag=0;
if(t<0)flag=1,t*=-1;
t*=100;
// cout<<t<<endl;
int t3=round(t);
// cout<<t3<<endl;
int t1=t3/100;int t2=(t3/10-t1*10)*6;
// cout<<t1<<" "<<t2<<endl;
if(flag)t*=-1;
if(t<0)x-=t1,y-=t2;
else x+=t1,y+=t2;
if(y<0)x--,y+=60;
if(y>=60)x++,y-=60;
if(x<0)x+=24;
if(x>=24)x-=24;
if(x<=9)printf("0%d:",x);
else printf("%d:",x);
if(y<=9)printf("0%d
",y);
else printf("%d
",y);
}
return 0;
}