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2019 ACM/TINA实验室10.12训练赛

由 跟風遠走 提交于 2019-12-01 06:08:53

A CodeForces 1238A Prime Subtraction

水题，除了差1都行。

#include<stdio.h>
#include<set>
#include<iostream>
#define mem(ss) memset(ss,0,sizeof(ss))
#define rep(d, s, t) for(int d=s;d<=t;d++)
#define rev(d, s, t) for(int d=s;d>=t;d--)
typedef long long ll;
typedef long double ld;
typedef double db;
const ll mod = 998244353;
const int N = 1e4 + 10;
#define io_opt ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;
ll T,x,y;
int main(){
    scanf("%lld",&T);
    while(T--){
        scanf("%lld%lld",&x,&y);
        if(x-y==1){
            printf("NO\n");
        }
        else{
            printf("YES\n");
        }
    }
    return 0;
}

B CodeForces 1238B Kill 'Em All

排序去重，从右向左炸。

#include<stdio.h>
#include<set>
#include<iostream>
#include<algorithm>
#define mem(ss) memset(ss,0,sizeof(ss))
#define rep(d, s, t) for(int d=s;d<=t;d++)
#define rev(d, s, t) for(int d=s;d>=t;d--)
typedef long long ll;
typedef long double ld;
typedef double db;
const ll mod = 998244353;
const int N = 1e4 + 10;
#define io_opt ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;
ll T,n,r;
ll a[100020];
ll cnt=0;
int main(){
    scanf("%lld",&T);
    while(T--){
        cnt=0;
        scanf("%lld%lld",&n,&r);
        for(int i=1;i<=n;i++){
            scanf("%lld",&a[i]);
        }
        sort(a+1,a+1+n);
        n=unique(a+1,a+1+n)-(a+1);
        for(int i=n;i>=1;i--){
            if(a[i]-cnt*r<=0){
                break;
            }
            cnt++;
        }
        printf("%lld\n",cnt);
    }
    return 0;
}

C CodeForces 1238C Standard Free2play

比赛想成dp了，还没补。

D CodeForces 540C Ice Cave

染色，分类讨论。

#include<stdio.h>
#include<set>
#include<iostream>
#include<algorithm>
#include<cstring>
#define mem(ss) memset(ss,0,sizeof(ss))
#define rep(d, s, t) for(int d=s;d<=t;d++)
#define rev(d, s, t) for(int d=s;d>=t;d--)
typedef long long ll;
typedef long double ld;
typedef double db;
const ll mod = 1000000007;
const int N = 1e4 + 10;
int f[5][5]={{0,1},{0,-1},{1,0},{-1,0}};
#define io_opt ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
const int INF = 1e8;
using namespace std;
int n,m;
int a[520][520];
string s;
int sx,sy,tx,ty,cnt=0;
void dfs(int cx,int cy){
    a[cx][cy]=cnt;
    for(int i=0;i<4;i++){
        int xx=cx+f[i][0],yy=cy+f[i][1];
        if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&a[xx][yy]==0){
            dfs(xx,yy);
        }
    }
}
int main() {
    io_opt;
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>s;
        for(int j=0;j<s.size();j++){
            if(s[j]=='X'){
                a[i][j+1]=-1;
            }
            else{
                a[i][j+1]=0;
            }
        }
    }
    cin>>sx>>sy>>tx>>ty;
    if(sx==tx&&sy==ty){
        for(int i=0;i<4;i++){
            int xx=sx+f[i][0],yy=sy+f[i][1];
            if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&a[xx][yy]==0){
                cout<<"YES\n";
                return 0;
            }
        }
        cout<<"NO\n";
        return 0;
    }
    int k=0;
    if(a[tx][ty]==-1){
        k=1;
    }
    a[sx][sy]=a[tx][ty]=0;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(a[i][j]==0){
                cnt++;
                dfs(i,j);
            }
        }
    }
    
    if(a[sx][sy]!=a[tx][ty]){
        cout<<"NO\n";
    }
    else{
        if(k==0){
            int tmp=0;
            for(int i=0;i<4;i++){
                int xx=tx+f[i][0],yy=ty+f[i][1];
                if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&a[xx][yy]!=-1){
                    tmp++;
                }
            }
            if(tmp>1){
                cout<<"YES\n";
            }
            else{
                cout<<"NO\n";
            }
        }
        else{
            cout<<"YES\n";
        }
    }
    return 0;
}

E CodeForces 545C Woodcutters

宽度为3的dp。

#include<stdio.h>
#include<set>
#include<iostream>
#include<algorithm>
#include<cstring>
#define mem(ss) memset(ss,0,sizeof(ss))
#define rep(d, s, t) for(int d=s;d<=t;d++)
#define rev(d, s, t) for(int d=s;d>=t;d--)
typedef long long ll;
typedef long double ld;
typedef double db;
const ll mod = 1000000007;
const int N = 1e4 + 10;
#define io_opt ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
const int INF = 1e8;
using namespace std;
int n,a[100020];
int f[100020][4];
struct E{
    int x,h;
}e[100020];
int main() {
    io_opt;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d%d",&e[i].x,&e[i].h);
    }
    e[0].x=-2e9-10;
    e[0].h=0;
    e[n+1].x=2e9+10;
    e[n+1].h=0;
    for(int i=1;i<=n;i++){
        f[i][0]=max(f[i-1][0],max(f[i-1][1],f[i-1][2]));
        if(e[i].x-e[i].h>e[i-1].x+e[i-1].h){
            f[i][1]=f[i][0]+1;
        }
        else if(e[i].x-e[i].h>e[i-1].x){
            f[i][1]=max(f[i][2],max(f[i-1][0],f[i-1][1])+1);
        }
        else{
            f[i][1]=f[i][0];
        }
        if(e[i].x+e[i].h<e[i+1].x){
            f[i][2]=f[i][0]+1;
        }
    }
    printf("%d\n",max(f[n][0],max(f[n][1],f[n][2])));
    return 0;
}

F CodeForces 1230A Dawid and Bags of Candies

之前组队打过，当时没注意1，3个也行，wa了好久。

#include<iostream>
 
using namespace std;
int a[10];
int sum, tot;
 
int main() {
    for (int i = 0; i < 4; i++) {
        cin >> a[i];
        sum += a[i];
    }
    if (a[0] + a[1] == a[2] + a[3] || a[0] + a[2] == a[1] + a[3] || a[0] + a[3] == a[1] + a[2] ||
        a[0] == a[1] + a[2] + a[3] || a[1] == a[0] + a[2] + a[3] || a[2] == a[0] + a[1] + a[3] ||
        a[3] == a[0] + a[1] + a[2]) {
        cout << "YES\n";
    } else {
        cout << "NO\n";
    }
    return 0;
}

G CodeForces 1230B Ania and Minimizing

贪心。

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cstring>
#include<stack>
 
#define io_opt ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define mem(arr) memset(arr,0,sizeof(arr))
#define rep(d, s, t) for(int d=s;d<=t;d++)
#define rev(d, s, t) for(int d=s;d>=t;d--)
typedef long long ll;
typedef long double ld;
typedef double db;
const ll mod = 998244353;
using namespace std;
 
ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
 
inline ll read() {
    ll data = 0;
    char ch = 0;
    while (ch < '0' || ch > '9') ch = getchar();
    while (ch >= '0' && ch <= '9') data = data * 10 + ch - '0', ch = getchar();
    return data;
}
 
const int N = 5e5 + 10;
ll n, k;
string s;
 
int main() {
    cin >> n >> k;
    cin >> s;
    if (k == 0) cout << s << endl;
    else if (n == 1) cout << '0'<<endl;
    else {
        if (s[0] != '1') s[0] = '1', --k;
        for (int i = 1; i < s.size(); i++) {
            if (k == 0) break;
            if (s[i] != '0') {
                s[i] = '0';
                --k;
            }
        }
        cout << s << endl;
    }
    return 0;
}

H CodeForces 1228B Filling the Grid

一开始以为那个是数量...
如果匹配，剩下的2的个数幂。

#include<stdio.h>
#include<set>
#include<iostream>
#include<algorithm>
#include<cstring>
#define mem(ss) memset(ss,0,sizeof(ss))
#define rep(d, s, t) for(int d=s;d<=t;d++)
#define rev(d, s, t) for(int d=s;d>=t;d--)
typedef long long ll;
typedef long double ld;
typedef double db;
const ll mod = 1000000007;
const int N = 1e4 + 10;
#define io_opt ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
const int INF = 1e8;
using namespace std;
int h,w;
int a[1020][1020];
int x,cnt;
ll lowspeed(ll a,ll b,ll p){
    ll cur=a,ans=0;
    while(b){
        if(b&1) ans=(ans+cur)%p;
        cur=(cur+cur)%p;
        b>>=1;
    }
    return ans%p;
}
ll speed(ll a,ll b,ll p){
    ll cur=a,ans=1;
    while(b){
        if(b&1) ans=lowspeed(ans,cur,p)%p;
        cur=lowspeed(cur,cur,p)%p;
        b>>=1;
    }
    return ans%p;
}
int main() {
    memset(a,-1,sizeof(a));
    scanf("%d%d",&h,&w);
    for(int i=1;i<=h;i++){
        scanf("%d",&x);
        for(int j=1;j<=x;j++){
            a[i][j]=1;
        }
        a[i][x+1]=0;
    }
    for(int i=1;i<=w;i++){
        scanf("%d",&x);
        for(int j=1;j<=x;j++){
            if(a[j][i]==0){
                printf("0\n");
                return 0;
            }
            a[j][i]=1;
        }
        if(a[x+1][i]==1){
            printf("0\n");
            return 0;
        }
        a[x+1][i]=0;
    }
    for(int i=1;i<=h;i++){
        for(int j=1;j<=w;j++){
            if(a[i][j]==-1) cnt++;
        }
    }
    ll ans=speed(2,cnt,mod);
    printf("%lld\n",ans);
    return 0;
}

I CodeForces 1228C Primes and Multiplication

题不难，就是麻烦，算就行。

#include<stdio.h>
#include<set>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#define mem(ss) memset(ss,0,sizeof(ss))
#define rep(d, s, t) for(int d=s;d<=t;d++)
#define rev(d, s, t) for(int d=s;d>=t;d--)
typedef long long ll;
typedef long double ld;
typedef double db;
const ll mod = 1e9+7;
const int N = 1e4 + 10;
#define io_opt ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
const int INF = 1e8;
using namespace std;
ll lowspeed(ll a,ll b,ll p){
    ll cur=a,ans=0;
    while(b){
        if(b&1) ans=(ans+cur)%p;
        cur=(cur+cur)%p;
        b>>=1;
    }
    return ans%p;
}
ll speed(ll a,ll b,ll p){
    ll cur=a,ans=1;
    while(b){
        if(b&1) ans=lowspeed(ans,cur,p)%p;
        cur=lowspeed(cur,cur,p)%p;
        b>>=1;
    }
    return ans%p;
}
const int MAXN=4e4;
bool ipr[MAXN+20];
int cnt,pri[MAXN/5];
void prime(){//∞£ Ω…∏∑®
    int N=sqrt(MAXN)+0.5,mul;
    memset(ipr,true,sizeof(ipr));
    ipr[1]=false;
    for(int i=2;i<=N;i++){
        if(ipr[i]==true){
            i==2?mul=1:mul=2;
            for(int j=i*i;j<=MAXN;j+=i*mul){
                ipr[j]=false;
            }
        }
    }
    for(int i=2;i<=MAXN;i++){
        if(ipr[i]==true){
            pri[++cnt]=i;
        }
    }
}
ll x,n,ans=1;
int inx[40020],ct=0;
int main() {
    prime();
    io_opt;
    cin>>x>>n;
    for(int i=1;i<=cnt&&x!=1;i++){
        if(x%pri[i]==0){
            inx[++ct]=pri[i];
        }
        while(x%pri[i]==0){
            x/=pri[i];
        }
    }
    if(x!=1){
        inx[++ct]=x;
    }
    ll tmp;
    for(int i=1;i<=ct;i++){
        tmp=n;
        while(tmp){
            tmp/=inx[i];
            ans=(ans*speed(inx[i],tmp,mod))%mod;
        }
    }
    cout<<ans<<endl;
    return 0;
}

J CodeForces 1210B Marcin and Training Camp

反向拓扑，用链表mle，用short的邻接矩阵存的。后来才知道有简单结论。

#include<stdio.h>
#include<set>
#include<iostream>
#include<algorithm>
#include<cstring>
#define mem(ss) memset(ss,0,sizeof(ss))
#define rep(d, s, t) for(int d=s;d<=t;d++)
#define rev(d, s, t) for(int d=s;d>=t;d--)
typedef long long ll;
typedef long double ld;
typedef double db;
const ll mod = 1000000007;
const int N = 1e4 + 10;
#define io_opt ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
const int INF = 1e8;
using namespace std;
int ag[7020];
short ae[7020][7020];
int cnt=0;
int val[7020];
ll ski[7020];
int n;
ll ans=0;
bool f[7020];
int du[7020];
int s;
void dfs(int cur){
    //printf("%d\n",cur);
    s--;
    f[cur]=true;
    ans-=val[cur];
    for(int i=1;i<=n;i++){
        if(ae[cur][i]){
            du[i]--;
        }
    }
    for(int i=1;i<=n;i++){
        if(!f[i]&&du[i]==s){
            dfs(i);
        }
    }
}

int main() {
    io_opt;
    cin>>n;
    s=n-1;
    for(int i=1;i<=n;i++){
        cin>>ski[i];
    }
    for(int i=1;i<=n;i++){
        cin>>val[i];
        ans+=val[i];
    }
    for(short i=1;i<=n;i++){
        for(short j=i+1;j<=n;j++){
            ll tmp=ski[i]^ski[j];
            if(tmp&ski[i]){
                //e[++cnt]=(E){i,j,g[i]};g[i]=cnt;
                //ae[++cnt]=(E){i,ag[j]};ag[j]=cnt;
                ae[j][i]=1;
                //printf("%d %d\n",i,j);
                du[i]++;
            }
            if(tmp&ski[j]){
                //e[++cnt]=(E){j,i,g[j]};g[j]=cnt;
                //ae[++cnt]=(E){j,ag[i]};ag[i]=cnt;
                ae[i][j]=1;
                //printf("%d %d\n",j,i);
                du[j]++;
            }
        }
    }
    //cout<<"???"<<endl;

    //cout<<"???"<<endl;
    //cout<<"!!!"<<du[5]<<endl;
    bool fg;
    do{
        fg=false;
        for(int i=1;i<=n;i++){
            //cout<<"du"<<du[i]<<endl;
            if(du[i]==s&&!f[i]){
                fg=true;
                dfs(i);
            }
        }
    }while(fg);
    //cout<<s<<endl;
    cout<<ans<<endl;
    return 0;
}

K CodeForces 1157D N Problems During K Days

L CodeForces 1157E Minimum Array

multiset，找不到就找后面一个，后面没了就找第一个。

#include<stdio.h>
#include<set>
#include<iostream>
#include<algorithm>
#include<cstring>
#define mem(ss) memset(ss,0,sizeof(ss))
#define rep(d, s, t) for(int d=s;d<=t;d++)
#define rev(d, s, t) for(int d=s;d>=t;d--)
typedef long long ll;
typedef long double ld;
typedef double db;
const ll mod = 1000000007;
const int N = 1e4 + 10;
int f[5][5]={{0,1},{0,-1},{1,0},{-1,0}};
#define io_opt ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
const int INF = 1e8;
using namespace std;
multiset<int>q;
int n;
int a[200020];
int b[200020];
int x;
int main() {
    io_opt;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    for(int j=1;j<=n;j++){
        cin>>x;
        q.insert(x);
    }
    for(int i=1;i<=n;i++){
        int tmp=n-a[i];
        auto cur=q.lower_bound(tmp);
        if(cur!=q.end()&&*cur==tmp){
            cout<<"0 ";
            q.erase(cur);
        }
        else{
            if(cur==q.end()){
                cout<<(a[i]+*q.begin())%n<<' ';
                q.erase(q.begin());
            }
            else{
                cout<<(a[i]+*cur)%n<<' ';
                q.erase(cur);
            }
        }
    }
    return 0;
}

M CodeForces 633H Fibonacci-ish II

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