A. Pineapple Incident
题解:
水题。。。注意没有t+1这种情形
代码:
#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define ll long long
using namespace std;
const int INF=1e9+10;
const int maxn=1000000+5;
ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
//-----------------------------------------------------------------------------
int main()
{
int t,s,x;
cin>>t>>s>>x;
if(x<t) cout<<"NO"<<endl;
else
{
int tmp=x-t,tmp1=x-t-1;
if(tmp==0||tmp%s==0||(tmp1%s==0&&tmp1!=0)) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}
B. Barnicle
题解:
将科学进制转换为10进制。。。。模拟题意就行。。注意有一个坑点
代码:
#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define ll long long
using namespace std;
const int INF=1e9+10;
const int maxn=1000000+5;
ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
//-----------------------------------------------------------------------------
int main()
{
string num;
cin>>num;
int pos1,pos2,d,b;
for(int i=0;i<num.length();i++)
{
if(num[i]==‘.‘) pos1=i;
if(num[i]==‘e‘) pos2=i;
}
b=0;
for(int i=pos2+1;i<num.length();i++) b=b*10+num[i]-‘0‘;
d=pos2-pos1-1;
if(b>=d)
{
int k=b-d,flag=0;
for(int i=0;i<pos2;i++)
{
if(i==pos1) continue;
if(num[i]==0&&flag==0) continue;
flag=1;
cout<<num[i];
}
for(int i=0;i<k;i++) cout<<‘0‘;
}
else
{
if(b==0)
{
if(d==1&&num[pos1+1]-‘0‘==0) for(int i=0;i<pos1;i++) cout<<num[i];
else for(int i=0;i<pos2;i++) cout<<num[i];
}
else
{
for(int i=0;i<=pos1+b;i++)
{
if(i==pos1) continue;
cout<<num[i];
}
cout<<num[pos1];
for(int i=pos1+b+1;i<pos2;i++) cout<<num[i];
}
}
return 0;
}
C. Lorenzo Von Matterhorn
题解:
使用map来保存，然后更新时从树向父节点更新。。。直到相等即可。求和也是如此
代码:
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define ll long long
const int INF=1e9+10;
const int maxn=1000000+5;
ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
//-----------------------------------------------------------------------------
map<pair<ll,ll>,ll> Edge;
void Change(ll a,ll b,ll c)
{
ll Max=max(a,b),Min=min(a,b);
while(Max!=Min)
{
while(Max>Min)
{
ll k=Max>>1;
Edge[make_pair(k,Max)]+=c;
Edge[make_pair(Max,k)]+=c;
Max=k;
}
while(Min>Max)
{
ll k=Min>>1;
Edge[make_pair(k,Min)]+=c;
Edge[make_pair(Min,k)]+=c;
Min=k;
}
}
}
ll solve(ll a,ll b)
{
ll sum=0;
ll Max=max(a,b),Min=min(a,b);
while(Max!=Min)
{
while(Max>Min)
{
ll k=Max>>1;
sum+=Edge[make_pair(Max,k)];
Max=k;
}
while(Min>Max)
{
ll k=Min>>1;
sum+=Edge[make_pair(Min,k)];
Min=k;
}
}
return sum;
}
int main()
{
int n;
ll d,a,b,c;
n=read();
for(int i=1;i<=n;i++)
{
d=read();
if(d==1)
{
a=read();b=read();c=read();
Change(a,b,c);
}
else
{
a=read();b=read();
cout<<solve(a,b)<<endl;
}
}
}
D. Puzzles
题解:
在dfs过程中。从父节点到子节点的过程中，每个子节点被访问的概率是相同的，求每个节点被访问的时间的期望
每个子节点被访问的概率是相同的等价于每个子节点都有50%的概率在其他的子节点之前被访问到，那么该节点的期望就等于父节点的期望+其余子节点的和/2;
先第一次求每个节点有多少节点。然后再求期望.两次dfs即可
代码:
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define ll long long
#define CLR(x) memset(x,0,sizeof x)
#define SZ(x) ((int)(x).size())
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
typedef pair<int,int> P;
const double eps=1e-9;
const int maxn=100100;
ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
//-----------------------------------------------------------------------------
vector<int> G[maxn];
double ans[maxn];
int sz[maxn];
double st[maxn],ct;
void dfs1(int s)
{
sz[s]=0;
for(int i=0;i<G[s].size();i++)
{
dfs1(G[s][i]);
sz[s]+=sz[G[s][i]];
}
sz[s]++;
}
void dfs2(int s)
{
for(int i=0;i<G[s].size();i++)
{
double sum=sz[s]-1-sz[G[s][i]];
st[G[s][i]]=st[s]+sum/2+1;
dfs2(G[s][i]);
}
}
int main()
{
int n,p;
cin>>n;
for(int i=2;i<=n;i++)
{
cin>>p;
G[p].pb(i);
}
ct=0;
st[1]=1;
dfs1(1);
dfs2(1);
for(int i=1;i<=n;i++) cout<<setprecision(10)<<st[i]<<" ";
return 0;
}
E. PLEASE
题解:
不怎么会做啊。。。数学方面的。
看的snowy_smile的博客(orz)
博客地址http://blog.csdn.net/snowy_smile/article/details/52052161
代码:
快速幂
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define ll long long
#define CLR(x) memset(x,0,sizeof x)
#define SZ(x) ((int)(x).size())
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define INF 2097152
typedef pair<int,int> P;
const double eps=1e-9;
const int maxn=1000100;
const int mod=1e9+7;
ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
//-----------------------------------------------------------------------------
int n;
ll mul(ll x,int p)
{
ll y=1;
while(p)
{
if(p&1) y=y*x%mod;
x=x*x%mod;
p>>=1;
}
return y;
}
ll inv(ll a)
{
return mul(a,mod-2);
}
void fast(int tim)
{
ll top=mul(2,tim);
if(tim&1)
{
top+=1;
if(top>=mod) top-=mod;
}
else
{
top-=1;
if(top<0) top+=mod;
}
top=top*inv(3)%mod;
cout<<top;
}
int main()
{
cin>>n;
ll tim=1;
for(int i=1;i<=n;i++)
{
ll x;cin>>x;
x%=(mod-1);
tim*=x;
tim%=(mod-1);
}
tim=(tim+mod-2)%(mod-1);
fast(tim);
cout<<"/";
cout<<mul(2,tim)<<endl;
return 0;
}
矩阵快速幂
/*
求4*n+1;
构造矩阵
4 0
1 1
初始矩阵
0 1
0 0
*/
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define ll long long
#define CLR(x) memset(x,0,sizeof x)
#define MC(x,y) memcpy(x,y,sizeof(x))
#define SZ(x) ((int)(x).size())
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define INF 2097152
typedef pair<int,int> P;
const double eps=1e-9;
const int maxn=1000100;
const int mod=1e9+7;
ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
//-----------------------------------------------------------------------------
int n;
const int G=2;
struct MX
{
int v[G][G];
void O(){CLR(v);}
void E(){CLR(v);for(int i=0;i<G;i++) v[i][i]=1;}
MX operator*(const MX &b) const
{
MX c;c.O();
for(int i=0;i<G;i++)
for(int j=0;j<G;j++)
for(int k=0;k<G;k++)
c.v[i][j]=(c.v[i][j]+(ll)v[i][k]*b.v[k][j])%mod;
return c;
}
MX operator+(const MX &b) const
{
MX c;c.O();
for(int i=0;i<G;i++)
for(int j=0;j<G;j++) c.v[i][j]=(v[i][j]+b.v[i][j])%mod;
return c;
}
MX operator^(ll p) const
{
MX y;y.E();
MX x;MC(x.v,v);
while(p)
{
if(p&1) y=y*x;
x=x*x;
p>>=1;
}
return y;
}
}a,b,c;
void tryit(int p)
{
b.O();
b.v[0][0]=4;
b.v[1][0]=b.v[1][1]=1;
a.O();
a.v[0][0]=0;
a.v[0][1]=1;
c=a*(b^(p/2));
ll ans=c.v[0][0];
if(p&1) ans=(ans*2+1)%mod;
cout<<ans;
}
ll mul(ll x,int p)
{
ll y=1;
while(p)
{
if(p&1) y=y*x%mod;
x=x*x%mod;
p>>=1;
}
return y;
}
int main()
{
cin>>n;
ll tim=1;
for(int i=1;i<=n;i++)
{
ll x;cin>>x;
x%=(mod-1);
tim*=x;
tim%=(mod-1);//这里用费马小定理进行加速，具体是这样的，要求2^n%mod,mod与2互质,那么2^(mod-1)%mod=1,求tim有多少个2即可
}
tim=(tim+mod-2)%(mod-1);//这里，因为在计算时最后用的是tim-1,这么写是为了避免tim=0时，减1变成负
tryit(tim);
cout<<"/";
cout<<mul(2,tim)<<endl;
return 0;
}
关于逆元的学习
http://www.cnblogs.com/linyujun/p/5194184.html
http://blog.csdn.net/hlyfalsy/article/details/38067021
http://blog.csdn.net/acdreamers/article/details/8220787
时间： 2024-10-08 17:24:11