今日SGU 5.5

SGU 114

题意：求一个点到其他点的距离总和最小，距离的定义是x轴距离乘以那个点的人数p

收获：带权中位数，按坐标排序，然后扫一遍，最后权值超过或等于总权值的一半时的那个点就是答案，证明暂无

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 1e5+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

//inv[1]=1;

//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;

struct P{

    ll x,p;

    bool operator <(const P c)const{

        return x<c.x;

    }

}a[maxn];

int main(){

    int n;

    scanf("%d",&n);

    ll sum = ;

    rep(i,,n) scanf("%lld%lld",&a[i].x,&a[i].p),sum+=a[i].p;

    sort(a,a+n);

    ll cnt = ;

    rep(i,,n) {

        cnt += a[i].p;

        if(*cnt>=sum) return printf("%lld\n",a[i].x),;

    }

    return ;

}

SGU 175

题意:定义一个phi函数，phi（1,2，.....，n） = phi(n,n-1,...,n/2+1) + phi(n/2,...,1);

问你初始在第p个位置的元素最后在哪

收获：二分去模拟，处理好对应的下标

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 1e5+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

//inv[1]=1;

//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;

int ans;

void phi(int l,int n,int p){

//    dd(l)dd(n)de(p)

    if(n==){

        ans = l;

        return ;

    }

    int mid = n/;

    if(mid>=p) phi(n-mid+l,mid,mid-p+);

    else phi(l,n-mid,n-p+);

}

int main(){

    int n,p;

    scanf("%d%d",&n,&p);

    phi(,n,p);

    printf("%d\n",ans);

    return ;

}

SGU 231

题意：问你有多少给<a,b>对，a<=b,且a，b为素数，且a+b<=n，且a+b也是素数，你会发现a一定是2（如果是其他的话，那么a+b就一定是偶数，所以a+b肯定不是素数），因为2是唯一一个偶数素数，且2是最小素数

收获：2是唯一一个偶数素数

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 1e5+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

//inv[1]=1;

//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;

const int N = 1e6+;

bool isPrime[N];

int prim[];

vector<int> ans;

int num = ;

int n;

void prime(){

    memset(isPrime,true,sizeof(isPrime));

    isPrime[] = isPrime[] = false;

    for(int i= ; i<=N ; i++){

        if( isPrime[i] ) prim[num++] = i;

        for(int j= ; j<num ; j++){

            if( i*prim[j]>N ) break;

            isPrime[ i*prim[j] ] = false;

            if( i%prim[j] ==  ) break;

        }

    }

//    de(num)

}

bool ok(int x,int y){

    if(x+y>n) return false;

    return isPrime[x+y];

}

int main(){

    prime();

    scanf("%d",&n);

    rep(i,,num){

        if(ok(,prim[i])) ans.pb(prim[i]);

    }

    printf("%d\n",sz(ans));

    rep(i,,sz(ans)) printf("2 %d\n",ans[i]);

    return ;

}

SGU 134

题意：求树的重心，重心的定义就是去除这点，然后分成的各个联通块的最大点数最小

收获：计算最大子树节点数

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = ;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

//inv[1]=1;

//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;

int n,u,v;

int minsubtree = inf;

int d[maxn];

vector<int> G[maxn];

vector<int> ans;

void dfs(int u,int p){

    int mn = ;

    d[u] = ;

    rep(i,,sz(G[u])){

        int v = G[u][i];

        if(v==p) continue;

        dfs(v,u);

        d[u] += d[v];

        mn = max(mn,d[v]);

    }

    mn = max(mn,n-d[u]);

//    dd(u)de(mn)

    minsubtree=min(mn,minsubtree);

}

void get_ans(int u,int p){

    int mn = ;

    d[u] = ;

    rep(i,,sz(G[u])){

        int v = G[u][i];

        if(v==p) continue;

        get_ans(v,u);

        d[u] += d[v];

        mn = max(mn,d[v]);

    }

    mn = max(mn,n-d[u]);

//    dd(u)de(mn)

    if(mn==minsubtree) ans.pb(u);

}

int main(){

    scanf("%d",&n);

    rep(i,,n) scanf("%d%d",&u,&v),G[u].pb(v),G[v].pb(u);

//    if(n==1) return printf("0 0\n1\n"),0;

    dfs(,-);mt(d,);get_ans(,-);

    sort(all(ans));

    printf("%d %d\n",minsubtree,sz(ans));

    rep(i,,sz(ans)) printf("%d%c",ans[i]," \n"[i+==sz(ans)]);

    return ;

}

//sgu对内存的要求真高

SGU 180

题意：求逆序对

收获：无

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = ;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

//inv[1]=1;

//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;

int a[maxn];

int tmp[maxn<<];

ll msort(int a[],int l,int r){

    if(l==r) return ;

    int i,j,k,m=(l+r)>>;

    ll t = msort(a,l,m) + msort(a,m+,r);

    for(i=l,j=m+,k=l;k<=r;++k){

        if(i<=m&&a[i]<=a[j]||j>r) tmp[k]=a[i++];

        else tmp[k]=a[j++],t+=m-i+;

    }

    memcpy(a+l,tmp+l,(r-l+)*sizeof(a[]));

    return t;

}

int main(){

    int n;

    scanf("%d",&n);

    rep(i,,n+) scanf("%d",a+i);

    printf("%lld\n",msort(a,,n));

    return ;

}

巴特西

今日SGU 5.5

最新文章

热门文章