AtCoder Regular Contest 068

AtCoder Regular Contest 068
题目链接： https://arc068.contest.atcoder.jp/
C.X: Yet Another Die Game（贪心+结论题）
D.Card Eater（贪心+结论题）
E.Snuke Line（思路+主席树）
我们考虑怎样的区间内不含k的倍数，显然就是0,k,2k,3k,…m之间的区间，即[1,k-1],[k+1,2k-1]…[xk+1,m]这些区间包含的区间。所以我们枚举k，然后枚举k的倍数，减去不含k的倍数的区间就是答案。怎么算这样的区间有多少个呢？类似于二维数点问题，给定一些点，询问一个矩形内的点数和。用主席树来做即可。复杂度O(mlogmlogm)。
F.Solitaire

C

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define ll long long#define inf 0x3f3f3f3f#define N 100010inline ll read(){    ll 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*10+ch-'0',ch=getchar();    return x*f;}int main(){//  freopen("a.in","r",stdin);    ll x=read(),ans=0;    ans+=x/11*2;x%=11;    if(x>6) ans+=2;    else if(x>0) ans+=1;    printf("%lld\n",ans);}

D

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;#define ll long long#define inf 0x3f3f3f3f#define N 100010inline 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*10+ch-'0',ch=getchar();    return x*f;}int n,cnt=0;bool flag[N];int main(){//  freopen("a.in","r",stdin);    n=read();    for(int i=1;i<=n;++i){        int x=read();if(!flag[x]) cnt++,flag[x]=1;    }printf("%d\n",(n-cnt)&1?cnt-1:cnt);    return 0;}

E

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define ll long long#define inf 0x3f3f3f3f#define N 100010inline 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*10+ch-'0',ch=getchar();    return x*f;}int n,m,rt[N],owo=0;struct node{    int lc,rc,x;}tree[N*3*20];struct point{    int x,y;}a[N*3];inline bool cmp(point x,point y){return x.x<y.x;}inline void add(int &p,int l,int r,int x){    tree[++owo]=tree[p];p=owo;tree[p].x++;    if(l==r) return;int mid=l+r>>1;    if(x<=mid) add(tree[p].lc,l,mid,x);    else add(tree[p].rc,mid+1,r,x);}inline int ask(int p1,int p2,int l,int r,int x,int y){    if(x<=l&&r<=y) return tree[p2].x-tree[p1].x;    int mid=l+r>>1,res=0;    if(x<=mid) res+=ask(tree[p1].lc,tree[p2].lc,l,mid,x,y);    if(y>mid) res+=ask(tree[p1].rc,tree[p2].rc,mid+1,r,x,y);    return res;}int main(){//  freopen("a.in","r",stdin);    n=read();m=read();    for(int i=1;i<=n;++i) a[i].x=read(),a[i].y=read();    sort(a+1,a+n+1,cmp);int now=1;    for(int i=1;i<=m;++i){        rt[i]=rt[i-1];        while(now<=n&&a[now].x==i){            add(rt[i],1,m,a[now].y);++now;        }    }printf("%d\n",n);    for(int i=2;i<=m;++i){        int ans=n,x=i;        for(x=i;x<=m;x+=i) ans-=ask(rt[x-i],rt[x-1],1,m,x-i+1,x-1);        if(x-i+1<=m) ans-=ask(rt[x-i],rt[m],1,m,x-i+1,m);printf("%d\n",ans);    }return 0;}