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ICM Technex 2017 and Codeforces Round #400 (Div. 1 + Div. 2, combined) 题解(A-E)

2017-02-24 10:55 726 查看

A A Serial Killer

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a
<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int 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 x*f;
}
int main()
{
//  freopen("A.in","r",stdin);
//  freopen(".out","w",stdout);
string s1,s2;
cin>>s1>>s2;
cout<<s1<<' '<<s2<<endl;
int n=read();
For(i,n) {
string s3,s4;
cin>>s3>>s4;
if (s3==s1) s1=s4;
else s2=s4;
cout<<s1<<' '<<s2<<endl;
}

return 0;
}


B Sherlock and his girlfriend

质数和合数染不同颜色就行

注意特判n=2,3的情况

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a
<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int 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 x*f;
}
#define MAXN (100000+10)
int p[MAXN],tot;
bool b[MAXN]={0};
void make_prime(int n)
{
tot=0;
Fork(i,2,n)
{
if (!b[i]) p[++tot]=i;
For(j,tot)
{
if (i*p[j]>n) break;
b[i*p[j]]=1;
if (i%p[j]==0) break;
}
}
}
int a[MAXN];
int main()
{
//  freopen("B.in","r",stdin);
//  freopen(".out","w",stdout);

int n=1+read();
make_prime(n+1);
if (n<=3) cout<<"1\n";else cout<<"2\n";
Fork(i,2,n-1) if (!b[i]) cout<<"1 ";else cout<<"2 ";
if (!b
) cout<<"1\n";else cout<<"2\n";

return 0;
}


C Molly’s Chemicals

经典题

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a
<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int 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 x*f;
}
#define MAXN (123456)
ll a[MAXN]={};
int n;
ll k;
vector<ll> v;
map<ll,int> S;
int main()
{
//  freopen("C.in","r",stdin);
//  freopen(".out","w",stdout);
cin>>n>>k;
For(i,n) cin>>a[i];

Fork(i,2,n) {
a[i]+=a[i-1];
}
ll p=1e16;
ll c=1;
v.pb(1);
if (k==-1) v.pb(-1);
else if (k!=1) {
while(c<=p) {
c*=k;
v.pb(c);
}
}
int sz=SI(v);
//  Rep(i,sz) cout<<v[i]<<' ';cout<<endl;
ll ans=0;
For(i,n) {
if (S.count(a[i-1])) S[a[i-1]]++;else S[a[i-1]]=1;
Rep(j,sz) {
ll pp=a[i]-v[j];
if (S.count(pp)) ans+=S[pp];
}
}
cout<<ans<<endl;
return 0;
}


D The Door Problem

经典题

以下解法均可解决问题:2-sat,并查集,dfs判连通,二分图染色。

判断一幅图有没有奇环的方法怎么有这么多

还有更狠的



PS:n,m敲反了FST好气啊

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a
<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int 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 x*f;
}
#define MAXN (223456)
class bingchaji
{
public:
int father[MAXN],n,cnt;
void mem(int _n)
{
n=cnt=_n;
For(i,n) father[i]=i;
}
int getfather(int x)
{
if (father[x]==x) return x;

return father[x]=getfather(father[x]);
}
void unite(int x,int y)
{
x=getfather(x);
y=getfather(y);
if (x^y) {
--cnt;
father[x]=y;
}
}
bool same(int x,int y)
{
return getfather(x)==getfather(y);
}
}S;
int a[MAXN];
int f[MAXN][3]={0};
int main()
{
//  freopen("D.in","r",stdin);
//  freopen(".out","w",stdout);
int n=read(),m=read();
S.mem(2*m);
For(i,n) a[i]=read();
For(i,m) {
int x=read();
For(j,x) {
int p=read();
f[p][++f[p][0]]=i;
}
}
n=m;
For(p,n) {
if (a[p]==0) S.unite(f[p][2],n+f[p][1]),S.unite(f[p][1],n+f[p][2]);
else S.unite(f[p][2],f[p][1]),S.unite(n+f[p][1],n+f[p][2]);
}
bool fl=0;
For(i,n) if (!a[i] &&S.same(i,n+i)) fl=1;
if(fl) puts("NO");else puts("YES");
return 0;
}


E The Holmes Children

f(i)=ϕ(i),g(i)=i

原题转变为求ϕ(ϕ(ϕ(...ϕ(n)...)))

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a
<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int 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 x*f;
}
ll phi(ll n) {
ll i,temp;
temp=n;
for(i=2;i*i<=n;i++)
{
if(n%i==0)
{
while(n%i==0) n=n/i;
temp=temp/i*(i-1);
}
if(n<i+1)
break;
}
if(n>1)
temp=temp/n*(n-1);
return temp;
}
ll n,k;
int main()
{
//  freopen("E.in","r",stdin);
//  freopen(".out","w",stdout);
cin>>n>>k;
for(ll i=1;i<=k;i+=2) {
n=phi(n);
if (n==1) break;
}
cout<<n%1000000007LL<<endl;
return 0;
}
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