11.2 模拟赛

菜的很的我又被踩了 耻辱#6

T1 meet

题目大意：

数轴上两个点x y 可以左移 右移1单位 或坐标*2 求最少步数

思路：

sb题 bfs就完事了

（我更sb 开始想错了以为有负数开小了空间）

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<vector>
 9 #include<map>
10 #define inf 2139062143
11 #define ll long long
12 #define MAXN 300100
13 using namespace std;
14 inline int read()
15 {
16     int x=0,f=1;char ch=getchar();
17     while(!isdigit(ch)) {if(ch=='-')f=-1;ch=getchar();}
18     while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
19     return x*f;
20 }
21 //yyc score=0
22 int n,k,q[MAXN<<3],l,r,vis[(MAXN<<1)+100];
23 int main()
24 {
25     freopen("meet.in","r",stdin);
26     freopen("meet.out","w",stdout);
27     n=read(),k=read();q[l=r=1]=n,vis[n]=1;int x;
28     if(n>=k) return printf("%d\n",n-k)&0;
29     while(l<=r)
30     {
31         x=q[l++];
32         if(x==k) break;
33         if(!vis[x+1]) vis[x+1]=vis[x]+1,q[++r]=x+1;
34         if(!vis[x-1]&&x-1>=0) vis[x-1]=vis[x]+1,q[++r]=x-1;
35         if(!vis[x*2]&&x*2<=4*k) vis[x*2]=vis[x]+1,q[++r]=x<<1;
36     }
37     printf("%d\n",vis[k]-1);
38 }
View Code

 

T2 sum

题目大意：

求一个数列的K阶前缀和数组

思路：

推一下式子发现是组合数

（然而我过于sb 没考虑n>k的情况）

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<vector>
 9 #include<map>
10 #define inf 2139062143
11 #define ll long long
12 #define MAXN 5100
13 #define MOD 1000000007
14 using namespace std;
15 inline int read()
16 {
17     int x=0,f=1;char ch=getchar();
18     while(!isdigit(ch)) {if(ch=='-')f=-1;ch=getchar();}
19     while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
20     return x*f;
21 }
22 //yyc score=0
23 ll n,k,a[MAXN],x[MAXN],ans[MAXN];
24 ll q_pow(ll bas,ll t)
25 {
26     ll res=1;
27     for(;t;t>>=1,(bas*=bas)%=MOD)
28         if(t&1) (res*=bas)%=MOD;
29     return res;
30 }
31 ll C(ll n,ll m)
32 {
33     if(n<m) return 0LL;
34     ll a=1LL,b=1LL;
35     while(m) (a*=n)%=MOD,(b*=m)%=MOD,n--,m--;
36     return a*q_pow(b,MOD-2)%MOD;
37 }
38 int main()
39 {
40     freopen("sum.in","r",stdin);
41     freopen("sum.out","w",stdout);
42     n=read(),k=read()-1;
43     for(int i=1;i<=n;i++) a[i]=read();
44     for(int i=x[0]=1;i<=n;i++) x[i]=C(k+i,i);
45     for(int i=1;i<=n;i++)
46         for(int t=0;t<i;t++) (ans[i]+=a[i-t]*x[t])%=MOD;
47     for(int i=1;i<=n;i++) printf("%lld ",ans[i]);
48 }
View Code

 

T3 xiaoqiao

题目大意：

极坐标（已被均分）染色

求被染过至少k次的面积

思路：

一看到题就快乐主席树结果不但写挂了还算错了空间会MLE

（非常不优秀的主席树代码得了95）

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<vector>
 9 #include<map>
10 #define inf 2139062143
11 #define ll long long
12 #define MAXN 200100
13 using namespace std;
14 inline int read()
15 {
16     int x=0,f=1;char ch=getchar();
17     while(!isdigit(ch)) {if(ch=='-')f=-1;ch=getchar();}
18     while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
19     return x*f;
20 }
21 //yyc score=0
22 int tot,n,m,lmt,rt[MAXN];
23 ll ans;
24 struct opt{int l,r,val;}q[MAXN];
25 struct node{int ls,rs,val;}tr[MAXN<<6];
26 bool cmp(const opt &a,const opt &b) {return a.val<b.val;}
27 void mdf(int &k,int kk,int l,int r,int a,int b)
28 {
29     if(!k) k=++tot,tr[k]=tr[kk];
30     if(l==a&&r==b) {tr[k].val++;return ;}
31     int mid=l+r>>1;
32     if(b<=mid) {tr[k].ls=0;mdf(tr[k].ls,tr[kk].ls,l,mid,a,b);}
33     else if(a>mid) {tr[k].rs=0;mdf(tr[k].rs,tr[kk].rs,mid+1,r,a,b);}
34     else {tr[k].ls=tr[k].rs=0;mdf(tr[k].ls,tr[kk].ls,l,mid,a,mid);mdf(tr[k].rs,tr[kk].rs,mid+1,r,mid+1,b);}
35 }
36 int v;
37 inline void queryv(int k,int l,int r,int x)
38 {
39     register int mid;
40     while(l!=r)
41     {
42         v+=tr[k].val,mid=l+r>>1;
43         if(x<=mid) r=mid,k=tr[k].ls;else l=mid+1,k=tr[k].rs;
44     }
45     v+=tr[k].val;
46 }
47 int main()
48 {
49     freopen("xiaoqiao.in","r",stdin);
50     freopen("xiaoqiao.out","w",stdout);
51     m=read(),n=read(),lmt=read();register int a,b,c;
52     for(register int i=1;i<=m;++i)
53     {
54         c=read(),a=read(),b=read();
55         if(a==n) a=-n;if(b==-n) b=n;
56         if(a<b) q[++tot].val=c,q[tot].l=a+n+1,q[tot].r=b+n;
57         else if(a+b==0) q[++tot].val=c,q[tot].l=1,q[tot].r=n;
58         else
59         {
60             q[++tot].val=c,q[tot].l=a+n+1,q[tot].r=n<<1;
61             q[++tot].val=c,q[tot].l=1,q[tot].r=b+n;
62         }
63     }
64     m=tot,n<<=1,tot=0;register int l,r,mid,res,Goal;
65     sort(q+1,q+m+1,cmp);
66     for(register int i=1;i<=m;++i)
67     {
68         //cout<<q[i].l<<" "<<q[i].r<<" "<<q[i].val<<endl;
69         mdf(rt[i],rt[i-1],1,n,q[i].l,q[i].r);
70     }
71     for(register int i=1;i<=n;++i)
72     {
73         l=1,r=m,res=v=0;queryv(rt[m],1,n,i);
74         if(v<lmt) continue;Goal=v-lmt+1;
75         while(l<=r)
76         {
77             mid=l+r>>1,v=0;queryv(rt[mid],1,n,i);
78             if(v>=Goal) r=mid-1,res=mid;
79             else l=mid+1;
80         }
81         ans+=(ll)q[res].val*q[res].val;
82         //cout<<i<<" "<<res<<" "<<q[res].val<<endl;
83     }
84     printf("%lld\n",ans);
85 }
View Code

正解应该是差分+扫描线

或使用平衡树/权值线段树 维护单点+ - 以及第k大 ($log$

（放上石神优秀的树状数组 $log^2$）

 1 #include<algorithm>
 2 #include<cmath>
 3 #include<cstdio>
 4 #include<cstdlib>
 5 #include<cstring>
 6 #include<ctime>
 7 #include<iomanip>
 8 #include<iostream>
 9 #include<map>
10 #include<stack>
11 #include<queue>
12 #include<vector>
13 #define rep(i,x,y) for(int i=(x);i<=(y);++i)
14 #define dwn(i,x,y) for(int i=(x);i>=(y);--i)
15 #define LL long long
16 #define maxn 200010
17 using namespace std;
18 int read()
19 {
20     int x=0,f=1;char ch=getchar();
21     while(!isdigit(ch)&&ch!='-')ch=getchar();
22     if(ch=='-')ch=getchar(),f=-1;
23     while(isdigit(ch))x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
24     return x*f;
25 }
26 void write(LL x)
27 {
28     if(x==0){putchar('0'),putchar('\n');return;}
29     if(x<0)putchar('-'),x=-x;
30     int f=0;char ch[20];
31     while(x)ch[++f]=x%10+'0',x/=10;
32     while(f)putchar(ch[f--]);
33     putchar('\n');
34     return;
35 }
36 int n,m,k,qr,qs,qt,tr[maxn];
37 int tpr;
38 LL ans;
39 int lt(int x){return x&(-x);}
40 void add(int x,int k){for(;x<=tpr;x+=lt(x))tr[x]+=k;return;}
41 int ask(int x){int k=0;for(;x;x-=lt(x))k+=tr[x];return k;}
42 vector<int>ad[maxn],de[maxn];
43 int gx(int x){return m-x;}
44 int nxt(int x){if(x>(m<<1))return x-(m<<1);if(x<1)return (m<<1)-x;return x;}
45 int main()
46 {
47     freopen("xiaoqiao.in","r",stdin);
48     freopen("xiaoqiao.out","w",stdout);
49     n=read(),m=read(),k=read();
50     rep(i,1,n)
51     {
52         qr=read(),qs=read(),qt=read();
53         qs=gx(qs),qt=gx(qt),qt=nxt(qt+1),qs=nxt(qs),swap(qs,qt);
54         if(qs<=qt)
55         {
56             ad[qs].push_back(qr),de[qt+1].push_back(qr);
57         }
58         else 
59         {
60             ad[1].push_back(qr),de[qt+1].push_back(qr);
61             ad[qs].push_back(qr),de[m*2+1].push_back(qr);
62         }
63         tpr=max(tpr,qr);
64     }tpr++;int li=m*2;
65     rep(i,1,li)
66     {
67         int lim=ad[i].size()-1;
68         rep(j,0,lim)add(1,1),add(ad[i][j]+1,-1);
69         lim=de[i].size()-1;
70         rep(j,0,lim)add(1,-1),add(de[i][j]+1,1);
71         int L=1,R=tpr,maxl=0;
72         while(L<=R)
73         {
74             int mid=(L+R>>1);
75             int tmp=ask(mid);
76             if(tmp>=k)maxl=max(maxl,mid),L=mid+1;
77             else R=mid-1;
78         }
79         ans+=(LL)maxl*(LL)maxl;
80     }
81     write(ans);
82     return 0;
83 }
84 /*
85 4 8 2
86 3 -8 8
87 3 -7 3
88 3 -5 5
89 3 7 6
90 */
View Code

 

posted @ 2018-11-02 18:58  jack_yyc  阅读(157)  评论(0)  编辑  收藏  举报