标签：
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
Codeforces Round #362 (Div. 2) ABCDE
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原文地址：http://www.cnblogs.com/byene/p/5745974.html
