Educational Codeforces Round 10

A:Gabriel and Caterpillar

题意：蜗牛爬树问题；值得一提的是在第n天如果恰好在天黑时爬到END，则恰好整除，不用再+1；

day = (End - Begin - day0)/(12*(up-down))+1;

#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <time.h>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <set>

#define c_false ios_base::sync_with_stdio(false); cin.tie(0)
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
#define zero_(x,y) memset(x , y , sizeof(x))
#define zero(x) memset(x , 0 , sizeof(x))
#define MAX(x) memset(x , 0x3f ,sizeof(x))
#define swa(x,y) {LL s;s=x;x=y;y=s;}
using namespace std ;
#define N 50005

const double PI = acos(-1.0);
typedef long long LL ;

int main(){
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    //ios_base::sync_with_stdio(false); cin.tie(0);
    int Begin, End, up, down;
    scanf("%d%d%d%d", &Begin, &End, &up, &down);
    int day0 = *up;
    int day;
    if(up-down <= ){
        if(End - Begin - day0 <=) day = ;
        else day = -;
    }else{
        if(End - Begin - day0 <=) day = ;
        else {
            if((End - Begin - day0)%(*(up-down)) == )
                day = (End - Begin - day0)/(*(up-down));
            else
                day = (End - Begin - day0)/(*(up-down))+;
        }
    }
    cout<<day;
    return ;
}

B:z-sort

题意：排序题；

#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <time.h>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <set>

#define c_false ios_base::sync_with_stdio(false); cin.tie(0)
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
#define zero_(x,y) memset(x , y , sizeof(x))
#define zero(x) memset(x , 0 , sizeof(x))
#define MAX(x) memset(x , 0x3f ,sizeof(x))
#define swa(x,y) {LL s;s=x;x=y;y=s;}
using namespace std ;
#define N 10005

const double PI = acos(-1.0);
typedef long long LL ;

int main(){
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    //ios_base::sync_with_stdio(false); cin.tie(0);
    int n, a[N], t[N];
    scanf("%d", &n);
    for(int i = ; i < n; i++){
        scanf("%d" ,&a[i]);
    }
    sort(a, a+n);
    int i = , j = ;
    for(; i < n; i+=, j++)
        t[i] = a[j];
    for(i = ; i < n; i+=, j++)
        t[i] = a[j];
    for(int k = ; k < n ; k++) printf("%d ", t[k]);
    return ;
}

C:Foe Pairs

ans=总区间数−非法的
经典离线；

所有区间如下：

(1,1),(2,2),(3,3),(4,4),(5,5)……(n,n)

(1,2),(2,3),(3,4),(4,5)……(n-1,n)

(1,3),(2,4),(3,5) ……(n-2,n)

(1,4),(2,5)……(n-3,n)

(1,5)……(n-4,n)

(1,n)

若给出非法对(2,4)

则非法区间是(1,4)与(2,4)以及其下方的所有区间；总数为:(n - 4+1)*(2);

要想在计算区间是不出现重复，则对于所有right值相同的非法对，只取left值最大的；

然后再对right值进行枚举：for：1->n;

ans += 1LL*(z[i]-pos) * (n - i +1);
pos = z[i];

也可以对left枚举；（此时的排序方法与上述不同）
时间复杂度O(nlogn)

#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <set>

#define c_false ios_base::sync_with_stdio(false); cin.tie(0)
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
#define zero_(x,y) memset(x , y , sizeof(x))
#define zero(x) memset(x , 0 , sizeof(x))
#define MAX(x) memset(x , 0x3f ,sizeof(x))
#define swa(x,y) {LL s;s=x;x=y;y=s;}
using namespace std ;
#define N 300005

const double PI = acos(-1.0);
typedef long long LL ;
int a[N], z[N], m, n;

int main(){
    //reopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    //ios_base::sync_with_stdio(false); cin.tie(0);
    scanf("%d%d", &n, &m);
    zero(a);zero(z);
    int x;
    for(int i = ; i <= n ; i++){
        scanf("%d", &x);
        a[x] = i;
    }
    int c,v;
    for(int i = ; i <= m; i++){
        scanf("%d%d", &c, &v);
        int t = max(a[c], a[v]);
        z[t] = max(z[t], min(a[c], a[v]));
    }
    LL ans = , pos = ;
    for(int i = ; i<= n ;i++){
        if(z[i] -pos >){
            ans += 1LL*(z[i]-pos) * (n - i +);
            pos = z[i];
        }
    }
    printf("%I64d\n", 1LL*n*(n+)/-ans);
    return ;
}

D:Nested Segments

思路: 离散化+树状数组

离散化：如果数据是2，10，1000，100000；则存储所需空间是1e5*4；而离散化存储只需要5*4，

　　　　数据只需要保证其位置不变即可，将2，10，1000，100000用1，2，3，4代替；

　　　　其位置关系，大小关系都不变，但是存储空间变小了；

　　　　这样更方便存储，使得使用树状数组成为可能；

剩下的就是树状数组了：按左区间排下序，然后累计右区间数内区间数；（按左区间从大到小累计，否则，嘿嘿）

时间复杂度：O(nlgn);

#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <set>

#define c_false ios_base::sync_with_stdio(false); cin.tie(0)
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
#define zero_(x,y) memset(x , y , sizeof(x))
#define zero(x) memset(x , 0 , sizeof(x))
#define MAX(x) memset(x , 0x3f ,sizeof(x))
#define swa(x,y) {LL s;s=x;x=y;y=s;}
using namespace std ;
#define N 200005
#define lowbit(k) k&(-k)
const double PI = acos(-1.0);
typedef long long LL ;

struct BIT{int l, r, id;};
int bit[N],ANS[N],n;
BIT num[N];
void update(int s, int k){
    for(int j = s; j <= n; j +=lowbit(j))
        bit[j] += k;
}
int query(int k){
    int ans = ;
    for(int i = k; i > ; i -=lowbit(i))
        ans += bit[i];
    return ans;
}
bool compL(BIT a, BIT b){return a.l<b.l;}
bool compR(BIT a, BIT b){return a.r<b.r;}
int main(){
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    //ios_base::sync_with_stdio(false); cin.tie(0);
    scanf("%d",&n);
    for(int i = ; i <= n; i++){
        scanf("%d%d", &num[i].l, &num[i].r);
        num[i].id = i;
    }
    sort(num+, num+n+, compR);
    //for(int i = 1; i <= n; i++) cout<<num[i].l<<"  "<<num[i].r<<"  "<<num[i].id<<endl;
    for(int i = ; i <= n; i++) num[i].r = i;
    sort(num+, num+n+, compL);
    //for(int i = 1; i <= n; i++) cout<<num[i].l<<"  "<<num[i].r<<"  "<<num[i].id<<endl;
    for(int i = n;i>=; i--){
        ANS[num[i].id] = query(num[i].r);
        update(num[i].r, );
    }
    for(int i = ; i <= n; i++) printf("%d\n", ANS[i]);
    return ;
}

E:Pursuit For Artifacts

思路：强连通分量

智商太低，无法理解，还是滚去看书去了；

题解连接：

http://www.cnblogs.com/Recoder/p/5323546.html

F:Ants on a Circle

思路：先做了poj 1852再说；