Atcoder Regular Contest 080 CDEF

C - 4-adjacent

轻量级分类讨论

#include <bits/stdc++.h>using namespace std;int n,cnt1,cnt2,cnt4,x;int main(){    scanf("%d",&n);    while (n--) {        scanf("%d",&x);        if (x%2!=0)            ++cnt1;        else if (x%4!=0)            ++cnt2;        else            ++cnt4;    }    if (cnt1<=cnt4)        puts("Yes");    else if (cnt1>cnt4+1)        puts("No");    else if (cnt1==cnt4+1&&!cnt2)        puts("Yes");    else        puts("No");    return 0;}

D - Grid Coloring

轻量级模拟

#include <bits/stdc++.h>using namespace std;int h,w,G[105][105],n;priority_queue<int> que;int main(){    scanf("%d%d%d",&h,&w,&n);    for (int i=0;i<n;++i) {        int t;        scanf("%d",&t);        while (t--)            que.push(i+1);    }    for (int i=0;i<h;++i) {        if (i&1) {            for (int j=0;j<w;++j) {                G[i][j]=que.top();                que.pop();            }        } else {            for (int j=w-1;j>=0;--j) {                G[i][j]=que.top();                que.pop();            }        }    }    for (int i=0;i<h;++i)        for (int j=0;j<w;++j)            printf("%d%c",G[i][j],j+1==w?'\n':' ');    return 0;}

E - Young Maids

倒推，维护区间偶数位置和奇数位置最小值

#include <bits/stdc++.h>using namespace std;typedef pair<int,int> pii;typedef pair<int,pii> pip;const int maxn=200005;int n,bit[maxn],a[maxn];pii odd[20][maxn],even[20][maxn];inline void init() {    memset(odd,0x3f,sizeof odd);    memset(even,0x3f,sizeof even);    for (int i=1;i<=n;++i)        if (i&1)            odd[0][i]=pii(a[i],i);        else            even[0][i]=pii(a[i],i);    for (int i=1;i<20;++i)        for (int j=1;j+(1<<i)-1<=n;++j) {            odd[i][j]=min(odd[i-1][j],odd[i-1][j+(1<<(i-1))]);            even[i][j]=min(even[i-1][j],even[i-1][j+(1<<(i-1))]);        }}inline pii oddmin(int l,int r) {    int x=0;    while ((1<<(x+1))<=r-l+1)        ++x;    return min(odd[x][l],odd[x][r-(1<<x)+1]);}inline pii evenmin(int l,int r) {    int x=0;    while ((1<<(x+1))<=r-l+1)        ++x;    return min(even[x][l],even[x][r-(1<<x)+1]);}priority_queue<pip,vector<pip>,greater<pip> > que;vector<int> res;int main(){    scanf("%d",&n);    for (int i=1;i<=n;++i)        scanf("%d",&a[i]);    init();    que.push(pip(oddmin(1,n).first,pii(1,n)));    while (!que.empty()) {        int l=que.top().second.first,r=que.top().second.second;        que.pop();        pii p1,p2;        if (l%2==0) {            p1=evenmin(l,r);            p2=oddmin(p1.second+1,r);        } else {            p1=oddmin(l,r);            p2=evenmin(p1.second+1,r);        }        res.push_back(p1.first);        res.push_back(p2.first);        if (p1.second-1>=l)            que.push(pip(l%2==1?oddmin(l,p1.second-1).first:evenmin(l,p1.second-1).first,pii(l,p1.second-1)));        if (p1.second+1<=p2.second-1)            que.push(pip((p1.second+1)%2==1?oddmin(p1.second+1,p2.second-1).first:evenmin(p1.second+1,p2.second-1).first,pii(p1.second+1,p2.second-1)));        if (p2.second+1<=r)            que.push(pip((p2.second+1)%2==1?oddmin(p2.second+1,r).first:evenmin(p2.second+1,r).first,pii(p2.second+1,r)));    }    for (int i=0;i<(int)res.size();++i)        printf("%d%c",res[i],i+1==(int)res.size()?'\n':' ');    return 0;}

F - Prime Flip

好题。用哥德巴赫猜想，转化成二分图最大匹配。

#include <bits/stdc++.h>using namespace std;const int maxn=105;int n,a[maxn],cnt[2],match[maxn<<1];bool used[maxn<<1];vector<int> G[maxn<<1],diff;set<int> vis;bool dfs(int v) {    used[v]=1;    for (int i=0;i<(int)G[v].size();++i) {        int u=G[v][i],w=match[u];        if (w==-1||(!used[w]&&dfs(w))) {            match[v]=u;            match[u]=v;            return true;        }    }    return false;}inline int bipartite_matching() {    int ret=0;    memset(match,-1,sizeof match);    for (int i=0;i<(int)diff.size();++i)        if (diff[i]%2==1&&match[i]==-1) {            memset(used,0,sizeof used);            ret+=dfs(i);        }    return ret;}bool check(int x) {    if (x<=2)        return false;    for (int i=2;i*i<=x;++i)        if (x%i==0)            return false;    return true;}int main(){    scanf("%d",&n);    for (int i=1;i<=n;++i) {        scanf("%d",&a[i]);        vis.insert(a[i]);    }    for (int i=1;i<=n;++i) {        if (vis.find(a[i]-1)==vis.end()) {            diff.push_back(a[i]);            ++cnt[a[i]%2];        }        if (vis.find(a[i]+1)==vis.end()) {            diff.push_back(a[i]+1);            ++cnt[(a[i]+1)%2];        }    }    for (int i=0;i<(int)diff.size();++i)        for (int j=i+1;j<(int)diff.size();++j)            if (check(abs(diff[i]-diff[j]))) {                G[i].push_back(j);                G[j].push_back(i);            }    int k=bipartite_matching();    printf("%d\n",k+2*((cnt[0]-k)/2+(cnt[1]-k)/2)+(cnt[0]-k)%2*3);    return 0;}