网赛的时候就是前三个小时过了后面五道,然后两个小时就没搞出啥了;账号密码都忘了,只好重新写一遍了;
hdu-5868
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define ll long long;
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('
');
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+2000;
const int maxn=2e4+50;
const double eps=1e-8;
int vis[maxn],b[maxn],cnt=0,prime[maxn];
inline void Init()
{
cnt=0;
for(int i=2;i<maxn;i++)
{
if(!vis[i])
{
for(int j=2*i;j<maxn;j+=i)vis[j]=1;
prime[++cnt]=i;
}
}
}
int phi(int x)
{
int ans=x;
for(int i=1;i<=cnt;i++)
{
if(x<prime[i])break;
if(x%prime[i]==0)
{
ans=ans/prime[i]*(prime[i]-1);
while(x%prime[i]==0)x/=prime[i];
}
}
if(x>1)ans=ans/x*(x-1);
return ans;
}
LL powmod(LL x,LL y)
{
LL s=1,base=x;
while(y)
{
if(y&1)s=s*base%mod;
base=base*base%mod;
y>>=1;
}
return s;
}
struct matrix
{
LL a[2][2];
};
matrix cal(matrix A,matrix B)
{
matrix C;
for(int i=0;i<2;i++)
{
for(int j=0;j<2;j++)
{
C.a[i][j]=0;
for(int k=0;k<2;k++)
{
C.a[i][j]+=A.a[i][k]*B.a[k][j];
C.a[i][j]%=mod;
}
}
}
return C;
}
LL pow_mod(int y)
{
matrix s,base;
s.a[0][0]=s.a[1][1]=1;s.a[0][1]=s.a[1][0]=0;
base.a[0][0]=base.a[0][1]=base.a[1][0]=1;base.a[1][1]=0;
while(y)
{
if(y&1)s=cal(s,base);
base=cal(base,base);
y>>=1;
}
return (s.a[0][0]*3+s.a[0][1])%mod;
}
LL solve(int x)
{
if(x==1)return 1;
else if(x==2)return 3;
else return pow_mod(x-2);
}
int main()
{
int n;
Init();
while(scanf("%d",&n)!=EOF)
{
if(n==1){printf("2
");continue;}
int le=(int)sqrt(n*1.0+0.5),num=0;
for(int i=1;i<=le;i++)
{
if(n%i==0)
{
b[++num]=n/i;
if(b[num]*b[num]==n)continue;
b[++num]=i;
}
}
LL ans=0;
for(int i=1;i<=num;i++)
{
ans=(ans+phi(n/b[i])*solve(b[i]))%mod;
}
ans=ans*powmod((LL)n,mod-2)%mod;
print(ans);
}
return 0;
}
/*
题意：
https://async.icpc-camp.org/d/546-2016
*/
hdu-5869
hdu-5870
hdu-5871
hdu-5872
hdu-5873
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define ll long long;
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('
');
}
const LL mod=1e4+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+2000;
const int maxn=2e4+50;
const double eps=1e-8;
int a[maxn];
int main()
{
int m;
while(scanf("%d",&m)!=EOF)
{
For(i,1,m)
{
int n,flag=1;
LL sum=0;
scanf("%d",&n);
For(i,1,n)scanf("%d",&a[i]);
sort(a+1,a+n+1);
for(int i=1;i<n;i++)
{
sum=sum+a[i];
if(sum<(LL)i*(i-1))flag=0;
}
sum=sum+a[n];
if(sum!=(LL)n*(n-1))flag=0;
if(flag)printf("T
");
else printf("F
");
}
}
return 0;
}
/*
题意：
赢了得两分,平局得一分,输了的分,问得分情况是否符合实际;
思路：
排序后看是否满足这种游戏的要求;读入挂不知怎么T了;
*/
hdu-5874
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define ll long long;
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('
');
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+2000;
const int maxn=2e4+50;
const double eps=1e-8;
int main()
{
LL n,m;
while(cin>>n>>m)
{
LL a=n/2,b=(n+1)/2;
if(a*b>m)printf("F
");
else printf("T
");
}
return 0;
}
/*
题意：
*/
hdu-5875
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define ll long long;
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('
');
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+2000;
const int maxn=1e5+50;
const double eps=1e-8;
int n,m,a[maxn],d[maxn][23];
inline void rmq()
{
for(int i=0;i<n;i++)d[i][0]=a[i];
for(int j=1;(1<<j)<=n;j++)
for(int i=0;i+(1<<j)<=n;i++)
d[i][j]=min(d[i][j-1],d[i+(1<<(j-1))][j-1]);
}
inline int query(int L,int R)
{
int k=0;
while((1<<(k+1))<=R-L+1)k++;
return min(d[L][k],d[R-(1<<k)+1][k]);
}
inline void solve(int u,int v)
{
int temp=a[u];
u++;
while(u<=v)
{
int l=u,r=v;
while(l<=r)
{
int mid=(l+r)>>1;
int t=query(u,mid);
if(t>temp)l=mid+1;
else r=mid-1;
}
if(l>v)break;
if(a[l])temp%=a[l];
if(temp==0)
{
printf("0
");
return ;
}
u=l+1;
}
printf("%d
",temp);
}
int main()
{
int t;
read(t);
while(t--)
{
read(n);
for(int i=0;i<n;i++)read(a[i]);
rmq();
read(m);
int l,r;
while(m--)
{
read(l);read(r);
l--;r--;
solve(l,r);
}
}
return 0;
}
/*
题意：
给一个数列,然后问[l,r]内这些数%得到多少;
思路：
比赛的时候队友用线段树A的,我现在写的是二分+rmq,就是让着里面的数递减才能改变它的结果;
*/
hdu-5876
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define ll long long;
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('
');
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+2000;
const int maxn=2e5+50;
const double eps=1e-8;
int n,m,ans[maxn];
pair<int,int>p;
map<pair<int,int>,int>mp;
queue<int>q,qu;
inline void makepair(int u,int v)
{
p.first=u;
p.second=v;
mp[p]=1;
}
int main()
{
int t;
read(t);
while(t--)
{
mp.clear();
read(n);read(m);
int u,v,s;
For(i,1,m)
{
read(u);read(v);
makepair(u,v);
makepair(v,u);
}
read(s);
while(!q.empty())q.pop();
while(!qu.empty())qu.pop();
For(i,1,n)
{
if(s==i)continue;
qu.push(i);
}
q.push(s);
ans[s]=0;
while(!q.empty())
{
int fr=q.front();
q.pop();
int siz=qu.size();
p.first=fr;
while(siz--)
{
int f=qu.front();
qu.pop();
p.second=f;
if(!mp[p])ans[f]=ans[fr]+1,q.push(f);
else qu.push(f);
}
}
int num=0;
For(i,1,n)
{
if(i==s)continue;
num++;
if(num<n-1)printf("%d ",ans[i]);
else printf("%d
",ans[i]);
}
}
return 0;
}
/*
题意：
求一个点多边少的补图的一个起点到其他所有点的最短距离;
思路：
补图里面的边就是原图里面没有的边,那么每次就判断两个点之间在原图中是否有边,主要是原图里面边较少可以这样判断;
*/
hdu-5877
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define ll long long;
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('
');
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+2000;
const int maxn=2e5+50;
const double eps=1e-8;
int a[maxn],b[maxn],c[maxn],cnt,sum[maxn],in[maxn],n;
LL k,ans;
vector<int>ve[maxn];
int lowbit(int x){return x&(-x);}
inline int query(int x)
{
int s=0;
while(x)
{
s+=sum[x];
x-=lowbit(x);
}
return s;
}
inline void update(int x,int num)
{
while(x<=cnt)
{
sum[x]+=num;
x+=lowbit(x);
}
}
inline int getpos(LL x)
{
int l=1,r=cnt;
while(l<=r)
{
int mid=(l+r)>>1;
if((LL)c[mid]>x)r=mid-1;
else l=mid+1;
}
return r;
}
void dfs(int cur)
{
int pos;
if(a[cur]==0)pos=cnt;
else pos=getpos(k/a[cur]);
ans=ans+query(pos);
int temp=getpos(a[cur]);
update(temp,1);
int len=ve[cur].size();
for(int i=0;i<len;i++)dfs(ve[cur][i]);
update(temp,-1);
}
int main()
{
int t;
read(t);
while(t--)
{
read(n);read(k);
For(i,0,n)
{
ve[i].clear();
sum[i]=in[i]=0;
}
For(i,1,n)read(a[i]),b[i]=a[i];
sort(b+1,b+n+1);
c[1]=b[1];cnt=1;ans=0;
For(i,2,n)
{
if(b[i]==b[i-1])continue;
c[++cnt]=b[i];
}
int u,v,root;
For(i,1,n-1)
{
read(u);read(v);
ve[u].push_back(v);
in[v]++;
}
For(i,1,n)
{
if(in[i])continue;
root=i;
break;
}
dfs(root);
print(ans);
}
return 0;
}
/*
题意：
求每个节点u以及和它的祖先节点v满足a[u]*a[v]<=k的对数和;
思路：
dfs的过程中将树状数组中保留的是它的所有祖先节点的信息,可以离散化后进行询问,
然后把这个点的信息更细到树状数组中,待它的所有节点都访问完了再删除,据说treap也可以做;
*/