比赛情况

bq. A题 Wrong Answer on test 2 , E题sb题没切。bqbqbq.

比赛总结

bq.

那就直接上题解吧！^-^

A

数位dp，分类讨论。

Talk is cheap.Show me the code.

B

我们把数值一样的数放在一起，扔进堆里。按数值从大到小处理就OK了。

注意值域比较大，用一下 \(STL\) 里面的 map。

Talk is cheap.Show me the code.

#include<bits/stdc++.h>
using namespace std;
inline int read() {
    int x=0,f=1; char ch=getchar();
    while(ch<'0' || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
    while(ch>='0'&&ch<='9') { x=(x<<3)+(x<<1)+(ch^48); ch=getchar(); }
    return x * f;
}
const int N = 2e5+7;
int n;
int a[N];
void work() {
    map<int,int> s;
    priority_queue<int> q;
    n = read(); int ans = 0;
    for(int i=1;i<=n;++i) {
        a[i] = read();
        if(s[a[i]]==0) {
            q.push(a[i]);
        }
        ++s[a[i]];
    }
    while(!q.empty()) {
        int now = q.top(); q.pop();
        if(now&1) continue;
        if(s[now/2]==0) q.push(now/2);
        s[now/2] += s[now]; ans++;
        //printf("ans += s[%d], s[now]=%d\n",now,s[now]);
    }
    printf("%d\n",ans);
}
int main()
{
    //freopen("a.out","w",stdout);
    int T = read();
    while(T--) work();
    return 0;
}

C

贪心题。

如果有 twone 则删去 'o'，否则删去中间这个，比如 one 删去 'n'。这样是对的。

然后模拟即可。

Talk is cheap.Show me the code.

#include<bits/stdc++.h>
using namespace std;
inline int read() {
    int x=0,f=1; char ch=getchar();
    while(ch<'0' || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
    while(ch>='0'&&ch<='9') { x=(x<<3)+(x<<1)+(ch^48); ch=getchar(); }
    return x * f;
}
void work() {
    string s; cin>>s; vector<int> Ans;
    for(int i=1;i<s.length();++i) {
        char a = s[i-1], b = s[i], c = s[i+1];
        if(a=='t' && b=='w' && c=='o') {
            if(s[i+2]=='n' && s[i+3]=='e') s[i+1] = '-', Ans.push_back(i+1)/*, i = i+3+1*/;
            else s[i] = '-', Ans.push_back(i)/*, i = i+1+1*/;
        } else if(a=='o' && b=='n' && c=='e') {
            Ans.push_back(i);
        }
    }
    printf("%d\n",(int)Ans.size());
    for(int i=0;i<Ans.size();++i) printf("%d ",Ans[i]+1);
    puts("");
}
int main()
{
    //freopen("a.out","w",stdout);
    int T = read();
    while(T--) work();
    return 0;
}

D

待补坑。

E

炒鸡简单

考虑一张下面这样的图：

答案不就是两个蓝色部分的大小相乘吗？（题目不考虑A，B两点）

这不就是Dfs的事吗qwq

（其实如果这两块蓝色的有边将他们连起来了就是 puts("0")）

Talk is cheap.Show me the code.

#include<bits/stdc++.h>
using namespace std;
inline int read() {
    int x=0,f=1; char ch=getchar();
    while(ch<'0' || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
    while(ch>='0'&&ch<='9') { x=(x<<3)+(x<<1)+(ch^48); ch=getchar(); }
    return x * f;
}
const int N = 2e5+7, M = 5e5+7;
int n,m,a,b,cnt,flag,tot;
int head[N],vis[N];
struct Edge {
    int next,to;
}edge[N<<1];
inline void add(int u,int v) {
    edge[++cnt] = (Edge)<%head[u],v%>;
    head[u] = cnt;
}
void Dfs(int u,int F) {
    flag |= (u==F); vis[u] = 1; ++tot;
    for(int i=head[u];i;i=edge[i].next) {
        int v = edge[i].to;
        if(!vis[v]) Dfs(v,F);
    }
}
void work() {
    memset(head, 0, sizeof(head)); cnt = flag = 0;
    n = read(), m = read(), a = read(), b = read();
    for(int i=1,u,v;i<=m;++i) {
        u = read(), v = read();
        add(u,v), add(v,u);
    }
    memset(vis, 0, sizeof(vis));
    vis[a] = 1; int n1 = 0, n2 = 0, nxt = 0;
    for(int i=head[a];i;i=edge[i].next) {
        int v = edge[i].to; tot = 0; Dfs(v,b);
        if(flag) {
            n1 = n - tot - 1; break;
        }
    }
    memset(vis, 0, sizeof(vis));
    vis[b] = 1; flag = 0;
    for(int i=head[b];i;i=edge[i].next) {
        int v = edge[i].to; tot = 0; Dfs(v,a);
        if(flag) {
            n2 = n - tot - 1; break;
        }
    }
    printf("%d\n",n1*n2);
}
int main()
{
    int T = read();
    while(T--) work();
    return 0;
}