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

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[n]<<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[n]<<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[n]) 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[n]<<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[n]<<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[n]<<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;}