Codeforces Round #447 (Div. 2)

Posted 2020-10-15 foreignbill

A

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

string st;

int main(){
    //freopen("in.txt","r",stdin);
    cin>>st;
    int cnt=0;
    for(int i=0;i<(int)st.size();i++)
        for(int j=i+1;j<(int)st.size();j++)
            for(int k=j+1;k<(int)st.size();k++)
                if(st[i]=='Q'&&st[j]=='A'&&st[k]=='Q')
                    cnt++;
    cout<<cnt;
    return 0;
}

B

猜结论。。。

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

ll n,m,k;

inline ll power(ll a,ll n,ll p){
    ll ret=1;ll now=a;
    while(n!=0){
        if(n&1)
            ret=ret*now%p;
        now=now*now%p;
        n>>=1;
    }
    return ret;
}

int main(){
    //freopen("in.txt","r",stdin);
    cin>>n>>m>>k;
    if(((n+m)&1)&&k==-1)
        puts("0");
    else {
        ll res = power(2,n-1,1000000007);
        res = power(res,m-1,1000000007);
        cout<<res;
    }
    return 0;
}

C

提供了一个假算法，惨遭hack。正解构造。
如果整个数列的最大公约数是最小的那个数字，那么可以构造出一个数列，只需要让每个数字中间夹着一个a[0] 就可以让区间长度大于等于二的区间的最大公约数为a[0]。

#include <stdio.h>
int n,a[1005];
int main(){
    scanf("%d",&n);
    for(int i=0;i<n;i++) scanf("%d",a+i);
    bool flg=1;
    for(int i=0;i<n;i++) if(a[i]%a[0]) flg=0;
    if(!flg) printf("-1");
    else {printf("%d\n",2*n);for(int i=0;i<n;i++) printf("%d %d ",a[i],a[0]);}
    return 0;
}

D

预处理出每个子树，每个点到该子树的根的距离，排个序。复杂度$O(n log n log n)$
然后对于每个询问，每次向上爬就行了。复杂度$O(m log n log n)$。

#include <bits/stdc++.h>
#define pb push_back
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 5;
vector<int> dis[maxn];
vector<ll> sum[maxn];
int n, m, A;
ll H;
int L[maxn];
int cnt;

template<typename T> inline void read(T &x) {
    x = 0; T f = 1; char ch; do {ch = getchar(); if (ch == '-')f = -1;} while (ch < '0' || ch > '9'); do x = x * 10 + ch - '0', ch = getchar(); while (ch <= '9' && ch >= '0'); x *= f;
}

template<typename A, typename B> inline void read(A&x, B&y) {read(x); read(y);}
template<typename A, typename B, typename C> inline void read(A&x, B&y, C&z) {read(x); read(y); read(z);}
template<typename A, typename B, typename C, typename D> inline void read(A&x, B&y, C&z, D&w) {read(x); read(y); read(z); read(w);}

void dfs(int id) {
    dis[id].pb(0);
    if (id * 2 > n) {
        return;
    } else if (id * 2 + 1 > n) {
        int l = id * 2;
        dfs(l);
        for (auto d : dis[l]) {
            dis[id].pb(d + L[l]);
        }
    } else {
        int l = id * 2, r = id * 2 + 1;
        dfs(l);
        for (auto d : dis[l]) {
            dis[id].pb(d + L[l]);
        }
        dfs(r);
        for (auto d : dis[r]) {
            dis[id].pb(d + L[r]);
        }
    }
}

ll get(int A, ll H) {
    int id = upper_bound(dis[A].begin(), dis[A].end(), H) - dis[A].begin();
    if (!id) return 0;
    return H * id - sum[A][id - 1];
}

int main() {
    //freopen("in.txt","r",stdin);
    read(n, m);
    for (int i = 2; i <= n; i++)
        read(L[i]);
    dfs(1);
    for (int i = 1; i <= n; i++)
        sort(dis[i].begin(), dis[i].end());
    for (int i = 1; i <= n; i++) {
        ll tmp = 0;
        for (int j = 0; j < (int)dis[i].size(); j++) {
            tmp += dis[i][j];
            sum[i].pb(tmp);
        }
    }
    ll xres;
    while (m--) {
        read(A, H);
        ll res = 0;
        res += xres = get(A, H);
        //cout<<"#"<<xres<<endl;
        int lst = A;
        int now = A / 2;
        while (now) {
            H -= L[lst];
            if (H < 0) break;
            res += H;
            if (lst & 1) {
                if (H >= L[now * 2]) {
                    res += xres = get(now * 2, H - L[now * 2]);
                    //cout<<"#"<<xres<<endl;
                }
            } else {
                if (H >= L[now * 2 + 1]) {
                    res += xres = get(now * 2 + 1, H - L[now * 2 + 1]);
                    //cout<<"#"<<xres<<endl;
                }
            }
            lst = now;
            now = now / 2;
        }
        printf("%I64d\n", res);
    }
    return 0;
}

E

首先，强连通分量中的每条边是都可以走"完"的。
对于边权w，可以二分一个最大的p，满足$w-1-2-...-p>=0$，此时的贡献为$w(p+1)-\frac{p(p+1)(p+2)}{6}$.
缩点之后做一次dag上的dp。

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e6+5;
const int maxm = 1e6+5;

template<typename T> inline void read(T &x){
x=0;T f=1;char ch;do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');do x=x*10+ch-'0',ch=getchar();while(ch<='9'&&ch>='0');x*=f;
}

template<typename A,typename B> inline void read(A&x,B&y){read(x);read(y);}
template<typename A,typename B,typename C> inline void read(A&x,B&y,C&z){read(x);read(y);read(z);}
template<typename A,typename B,typename C,typename D> inline void read(A&x,B&y,C&z,D&w){read(x);read(y);read(z);read(w);}

struct node{
    int v;ll w;
    node(){}
    node(int _v,ll _w):v(_v),w(_w){}
};
vector<node> e[maxn];
int Low[maxn],DFN[maxn],Stack[maxn],Belong[maxn];
int num[maxn];
int Index,top,n,m,x,y;
ll z;
int scc;
bool Instack[maxn];
ll score[maxn];
vector<int> scce[maxn];
ll dp[maxn];

void Tarjan(int u){
    int v;
    Low[u]=DFN[u]=++Index;
    Stack[top++]=u;
    Instack[u]=true;
    for(auto enxt:e[u]){
        v=enxt.v;
        if(!DFN[v]){
            Tarjan(v);
            Low[u]=min(Low[u],Low[v]);
        } else if(Instack[v] && Low[u]>DFN[v])
            Low[u]=DFN[v];
    }
    if(Low[u]==DFN[u]){
        scc++;
        do{
            v=Stack[--top];
            Instack[v]=false;
            Belong[v]=scc;
            num[scc]++;
        }while(v!=u);
    }
}
int indeg[maxn],outdeg[maxn];

ll calc(ll w){
    ll l=0,r=1e6+5,rt;
    while(l<=r){
        ll mid=(l+r)/2;
        if(mid*(mid+1)/2<=w)
            l=(rt=mid)+1;
        else r=mid-1;
    }
    ll ret=w*(rt+1ll)-rt*(rt+1ll)*(rt+2ll)/6ll;
    return ret;
}

inline void solve(int n){
    memset(DFN,0,sizeof DFN);
    memset(num,0,sizeof num);
    memset(Instack,0,sizeof Instack);
    Index=scc=top=0;
    for(int i=1;i<=n;i++) if(!DFN[i]) Tarjan(i);
    memset(indeg,0,sizeof indeg);
    memset(outdeg,0,sizeof outdeg);
    for(int u=1;u<=n;u++)
        scce[Belong[u]].push_back(u);
    for(int u=1;u<=n;u++)
        for(auto enxt:e[u]){
            int v=enxt.v;
            if(Belong[u]!=Belong[v]){
                outdeg[Belong[u]]++;
                indeg[Belong[v]]++;
            } else {
                score[Belong[u]]+=calc(enxt.w);
            }
        }
}

ll find(int nowscc){
    if(dp[nowscc]!=-1) return dp[nowscc];
    ll ret = 0;
    for(auto u:scce[nowscc])
        for(auto enxt:e[u]){
            int v=enxt.v;
            if(Belong[u]==Belong[v]) continue;
            ll w=enxt.w;
            ret = max(ret,find(Belong[v])+w);
        }
    dp[nowscc]=(ret+score[nowscc]);
    return dp[nowscc];
}

int main(){
    //freopen("in.txt","r",stdin);
    read(n,m);
    for(int i=1;i<=m;i++){
        read(x,y,z);
        e[x].push_back(node(y,z));
    }
    solve(n);
    // printf("%d\n",Belong[0]);
    // for(int i=1;i<=scc;i++){
    //  for(auto ep:scce[i])
    //      printf("%d ",ep);
    //  puts("");
    // }
    memset(dp,-1,sizeof dp);
    int start;
    read(start);
    ll res = find(Belong[start]);
    cout<<res;
    return 0;
}