2019牛客网多校赛（第一场）题解

官方题解：https://ac.nowcoder.com/discuss/208642?type=101&order=0&pos=9&page=1

 

A ： Equivalent Prefixes

题目链接 ： https://ac.nowcoder.com/acm/contest/881/A

 

题解： 

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

 

B Integration 

题目链接：https://ac.nowcoder.com/acm/contest/881/B

题目描述：

       

 

题解： 

 

思路： https://blog.csdn.net/qq_41730082/article/details/96443875

c++代码：

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include

 

C :   Euclidean Distance

题目描述 ：

题解： 

 

C++ 代码：

#include 
using namespace std;
#define debug(a) cout << #a": " << a << endl;
#define mst(a, b) memset(a, b, sizeof(a))
#define ALL(x) x.begin(), x.end()
#define lc rt << 1
#define rc rt << 1 | 1
#define X first
#define Y second
inline int lowbit(int x) { return x & -x; }
typedef long long LL;
typedef unsigned long long ULL;
typedef double db;
typedef pair pii;
const int N = 1e4 + 10;
const int M = 1e5 + 10;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
const int mod = 1e9 + 7;
 
LL a[N];
LL pre[N];
 
int main() {
#ifdef purple_bro
    freopen("in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
 
    int n, m;
 
    for (;cin >> n >> m;) {
        for (int i = 1; i <= n; i++)
            cin >> a[i];
 
        sort(a + 1, a + 1 + n, [](int a, int b) -> bool {
                return a > b;
             });
 
        pre[0] = -m;
 
        for (int i = 1; i <= n; i++)
            pre[i] = pre[i - 1] + a[i];
 
        int now = n;
 
        for (int i = 1; i < n; i++) if (pre[i] > a[i + 1] * i) {
            now = i;
            break;
        }
 
        LL t = pre[now] * pre[now];
        LL d = now;
        for (int i = now + 1; i <= n; i++)
            t += a[i] * a[i] * d;
        d *= m * m;
        LL gcd = __gcd(t, d);
        t /= gcd; d /= gcd;
        if (d == 1 || !t)
            cout << t << "\n";
        else
            cout << t << "/" << d << "\n";
    }
 
    return 0;
}

 

D:  Parity of Tuples

 

 

 

#include
#include
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
ll i,j,u,vv,n,m,k,a[11],kp,dp[1048576],inv2,v[1024],l[1024],pk,pt,cnt[1048576],pm;
ll lowbit(ll p){return p&-p;}
ll M(ll p)
{
    if(p<0)return p+mod;
    if(p>=mod)return p-mod;
    return p;
}
ll pow(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1)ans=ans*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ans;
}
int main()
{
    inv2=pow(2LL,mod-2LL);
    cnt[0]=1;
    for(i=1;i<(1<<20);i++)cnt[i]=-cnt[i-lowbit(i)];
    while(scanf("%lld%lld%lld",&n,&m,&k)!=EOF)
    {
        //init();
        for(i=0;i<(1<

 

E : ABBA

样例：

20 36
40 69
100 1000
1000 200
200 99
99 1
 
 
1 2
0 1dp1
0 2dp1
1 0dp1
1 1dp2
1 2dp3
1 3dp4
2 1dp3
2 2dp6
2 3dp10
3 1dp4
3 2dp10
3 3dp20
 
 
20 36
677063087
40 69
357970915
100 1000
651709122
1000 200
946791934
200 99
568337085
99 1
554042848

 

 

#include 
#include 
using namespace std;
#define MOD 1000000007
int dp[3000][3000];
 
int main()
{
    int n, m;
    while(cin >> n >> m)
    {
        for(int i = 0; i <= n + m; ++i)
        {
            for(int j = 0; j <= n + m; ++j)
            {
                dp[i][j] = 0;
            }
        }
        dp[0][0] = 1;
        for(int i = 1; i <= m; ++i)
        {
            dp[0][i] = 1;
        }
        for(int i = 1; i <= n; ++i)
        {
            dp[i][0] = 1;
        }
 
        for(int i = 1; i <= n + m; ++i)
        {
            for(int j = 1; j <= n + m; ++j)
            {
                if(i <= n + j&&j <= m + i)
                    dp[i][j] = (dp[i - 1][j] + dp[i][j - 1])%MOD;
            }
        }
        cout << dp[n + m][n + m]<

 

F  : Random Point in Trangle

 

#include 
  
using namespace std;
  
typedef long long ll;
  
#ifdef iq
  mt19937 rnd(228);
#else
  mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
#endif
  
int main() {
#ifdef iq
  freopen("a.in", "r", stdin);
#endif
  ios::sync_with_stdio(0);
  cin.tie(0);
  int x1, y1, x2, y2, x3, y3;
  while (cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3) {
    ll a = x2 - x1, b = y2 - y1;
    ll c = x3 - x1, d = y3 - y1;
    ll ans = abs(a * d - b * c);
    cout << ans * 11 << '\n';
  }
}

 

G : Substrings 2

 

 

#include
#define rep(i,x,y) for(auto i=(x);i<=(y);++i)
#define dep(i,x,y) for(auto i=(x);i>=(y);--i)
#define see(x) (cerr<<(#x)<<'='<<(x)<pii;
const int N=5e4+10;
const int M=300;
const ll mo=998244353;
const ll D=18113219;
const ll DL=181321;
ll pw[M],s[N][M],nw[N][M];int n,m,b[N],l[N],r[N];
inline bool cmp(int x,int y){
    rep(i,0,m)if(s[x][i]!=s[y][i]){
        rep(j,0,m){int p=m*i+j;
            if(x+p>n)return 1;
            if(y+p>n)return 0;
            int X=l[x+p]Y)return 0;
        }
    }return 0;
}
void sol(){
    if(scanf("%d",&n)==EOF)exit(0);
    mapmp;m=sqrt(n)+1;
    rep(i,1,n){int x;
        scanf("%d",&x);
        if(mp[x])r[mp[x]]=i;
        l[i]=mp[x];r[i]=0;mp[x]=i;
    }
    rep(i,0,m)nw[n][i]=s[n][i]=0;
    s[n][0]=DL;
    dep(i,n-1,1){ll ls=DL;bool lf=0;
        rep(j,0,(n-i)/m){
            int p=i+(j+1)*m,c;
            nw[i][j]=(nw[i+1][j]*D+lf)%mo;lf=0;
            if(r[i]&&(r[i]-i)/m==j)(nw[i][j]+=pw[r[i]-(i+j*m)])%=mo;
            if(p<=n&&l[p]>i){
                lf=1;(nw[i][j]+=mo-pw[m])%=mo;
            }
            if(p<=n){
                if(l[p]<=i)c=DL;else c=DL+l[p]-i;
            }else c=0;
            s[i][j]=(s[i+1][j]*D+mo-c*pw[m]%mo+ls+nw[i][j])%mo;ls=c;
        }
    }
    rep(i,1,n)b[i]=i;
    sort(b+1,b+n+1,cmp);
    ll ans=n-b[1]+1;
    rep(i,2,n){int lcp=0;
        rep(j,0,m)if(s[b[i-1]][j]!=s[b[i]][j]){
            rep(k,0,m){int p=m*j+k;
                if(max(b[i-1],b[i])+p>n||
                (l[b[i-1]+p]

 

 

H:  XOR 

具体题面：https://blog.csdn.net/u013534123/article/details/96482572

 

 

#include
using namespace std;
typedef long long ll;
 
const int MN=63;
const int mod=1000000007;
 
ll num[100005];
 
 
struct LinearBase {
    ll base[MN];
 
    bool flag;//该线性基能否表示0
    int cnt;
 
    void Copy(LinearBase b) {
        cnt=b.cnt;
        flag=b.flag;
        memcpy(base,b.base,sizeof(base));
    }
 
    void Clear() {
        cnt=0;
        flag=false;
        memset(base,0,sizeof( base));
    }
 
    //尝试向线性基中插入一个值
    void Insert(ll x) {
        for(int i=MN; ~i; i--)
            if(x&(1ll<>=1;
    }
    return res;
}
 
ll ans,p2;
 
vectorB1ID;
 
int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
    //freopen("Yinku.out", "w", stdout);
#endif // Yinku
    int n;
    while(~scanf("%d", &n)) {
 
        //cout<<"0.\n";
        for(int i=1; i<=n; i++) {
            scanf("%lld",&num[i]);
        }
 
        B1.Clear();
        B2.Clear();
        B1ID.clear();
 
        for(int i=1; i<=n; i++) {
            if(B1.Check(num[i])) {
                B2.Insert(num[i]);
            } else {
                B1.Insert(num[i]);
                B1ID.push_back(i);
            }
        }
 
        ans=0;
        if(n!=B1.cnt) {
            p2=qpow(2,n-B1.cnt-1);
            ans+=p2*(n-B1.cnt)%mod;
        }
 
        for(ll i:B1ID) {
            B3.Copy(B2);
            for(ll j:B1ID) {
                if(i!=j) {
                    B3.Insert(num[j]);
                }
            }
            if(B3.Check(num[i])) {
                //num[i]能被其他数表示
                ans=(ans+p2)%mod;
            }
        }
 
        printf("%lld\n", ans);
    }
}

 

 

I题

Points  Division

#include
using namespace std;
#define ll long long
const int N=100010;
const ll inf=1000000000000007ll;
 
#define ls (x<<1)
#define rs (x<<1|1)
#define mid (l+r>>1)
ll mx[N<<2],ad[N<<2];
void pushup(int x){
    mx[x]=max(mx[ls],mx[rs]);
}
void pushdown(int x){
    ad[ls]+=ad[x];ad[rs]+=ad[x];
    mx[ls]+=ad[x];mx[rs]+=ad[x];
    ad[x]=0;
}
void build(int x,int l,int r){
    mx[x]=0;ad[x]=0;
    if(l==r)return ;
    build(ls,l,mid);build(rs,mid+1,r);
}
void add(int x,int l,int r,int L,int R,ll p){
    if(L>r||l>R)return ;
    if(L<=l&&r<=R){
        mx[x]+=p;ad[x]+=p;
        return ;
    }
    pushdown(x);
    add(ls,l,mid,L,R,p);
    add(rs,mid+1,r,L,R,p);
    pushup(x);
}
void change(int x,int l,int r,int pos,ll v){
    if(posr)return ;
    if(l==r){
        mx[x]=max(mx[x],v);
        return ;
    }
    pushdown(x);
    change(ls,l,mid,pos,v);
    change(rs,mid+1,r,pos,v);
    pushup(x);
}
ll query(int x,int l,int r,int L,int R){
    if(L>r||l>R)return -inf;
    if(L<=l&&r<=R){
        return mx[x];
    }
    pushdown(x);
    return max(query(ls,l,mid,L,R),query(rs,mid+1,r,L,R));
}
int n;
struct E{
    int x,y,a,b;
}a[N];
bool comp(E A,E B){
    if(A.x==B.x)return A.y>B.y;
    return A.x

 

J题

Fraction Comparision

 

#include 
#define For(i,x,y) for(int i=x; i
#define INF 0x3f3f3f3f
#define ll long long
#define se second
#define fi first
#define maxn 100005
#define endl '\n'
using namespace std;
ll gcd(ll a ,ll b){
    if(b) return gcd(b,a%b);
    return a;
}
int main(){
    ll x,y,a,b;
    while(cin>>x>>a>>y>>b){
        if(x*b==y*a) cout<<'='<y/b) { cout<<'>'<y*a) cout<<'>'<