Codeforces Round #368 (Div. 2) ABCD

A. Brain‘s Photos

题解:

水得不要不要的

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 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
typedef pair<int,int> P;
const double eps=1e-9;
const int maxn=200;
const int mod=1e9+7;
const int INF=1e9;

string p;
set<string> s;

int main(){
    s.insert("C");
    s.insert("M");
    s.insert("Y");
    int n,m,flag=0;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++){
        cin>>p;
        if(flag) continue;
        if(s.count(p)) flag=1;
    }
    if(!flag) cout<<"#Black&White"<<endl;
    else      cout<<"#Color"<<endl;
    return 0;
}

B.Chris and Magic Square

题解:

英语阅读题，看懂题意就很水了

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 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
typedef pair<int,int> P;
const double eps=1e-9;
const int maxn=1e5+10;
const int mod=1e9+7;
const int INF=1e9+10;

int v[maxn],u[maxn],w[maxn];
set<int> s;

int main(){
    int n,m,k,tmp;
    cin>>n>>m>>k;
    for(int i=1;i<=m;i++) cin>>v[i]>>u[i]>>w[i];
    for(int i=1;i<=k;i++){
        cin>>tmp;
        s.insert(tmp);
    }
    int ans=INF;
    for(int i=1;i<=m;i++){
        if((s.count(v[i])&&!s.count(u[i]))||(!s.count(v[i])&&s.count(u[i])))
        ans=min(w[i],ans);
    }
    if(ans==INF) cout<<-1<<endl;
    else       cout<<ans<<endl;
}

C. Pythagorean Triples

题解:

百度一下勾股数就行了。。。

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 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
typedef pair<int,int> P;
const double eps=1e-9;
const int maxn=200;
const int mod=1e9+7;
const int INF=1e9;

LL n,m,a,b,c;

int main(){
    cin>>n;
    if(n<3) cout<<-1<<endl;
    else{
        if(n&1){
            b=(n*n-1*1LL)/2*1LL;
            c=(n*n+1*1LL)/2*1LL;
        }
        else{
            b=(n*n/2-2*1LL)/2*1LL;
            c=(n*n/2+2*1LL)/2*1LL;
        }
        cout<<b<<" "<<c<<endl;
    }
    return 0;
}

D. Persistent Bookcase

题解:

考验码力的题目。。用上bitset，然后模拟就行了。

注意到一点，4种操作，更改的都是一行的值

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 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
typedef pair<int,int> P;
const double eps=1e-9;
const int N=1e5+10;
const int M=1e3+10;
const int mod=1e9+7;
const int INF=1e9+10;

bitset<M> bs[N],tmp;
int que[N][M],ans[N],cnt;

int main(){
    int n,m,q,a,b,c;
    cnt=0;
    scanf("%d%d%d",&n,&m,&q);
    for(int i=1;i<=m;i++) tmp[i]=1;
    for(int i=1;i<=q;i++){
        scanf("%d%d",&a,&b);
        if(a==4){
            for(int j=1;j<=n;j++) que[i][j]=que[b][j];
            ans[i]=ans[b];
            continue;
        }
        for(int j=1;j<=n;j++) que[i][j]=que[i-1][j];
        ans[i]=ans[i-1];
        int id=que[i-1][b];
        if(a==3){
            bs[++cnt]=bs[id];
            ans[i]-=bs[cnt].count();
            bs[cnt]^=tmp;
            ans[i]+=bs[cnt].count();
            que[i][b]=cnt;
        }
        if(a==2){
            scanf("%d",&c);
            if(bs[id][c]==0) continue;
            bs[++cnt]=bs[id];
            bs[cnt][c]=0;
            ans[i]--;
            que[i][b]=cnt;
        }
        if(a==1){
            scanf("%d",&c);
            if(bs[id][c]==1) continue;
            bs[++cnt]=bs[id];
            bs[cnt][c]=1;
            ans[i]++;
            que[i][b]=cnt;
        }
    }
    for(int i=1;i<=q;i++) printf("%d\n",ans[i]);
    return 0;
}