CodeForces Round #514 (div2)

A：Cashier

题意：问可以休息多少次。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1e5 + 100;
int n, L, a;
int t[N], l[N];
int main(){
    scanf("%d%d%d", &n, &L, &a);
    for(int i = 1; i <= n; ++i)
        scanf("%d%d", &t[i], &l[i]);
    t[n+1] = L;
    t[0] = l[0] = 0;
    int ans = 0;
    for(int i = 0; i <= n; ++i){
        int dif = t[i+1] - t[i] - l[i];
        ans += dif/a;
    }
    cout << ans << endl;
    return 0;
}

B：B - Forgery

题意：一次能给边上一圈填上'#'，问这个图形合法不合法。填的时候不能越界。

直接去check行不行就好啦。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1e3 + 100;
char s[N][N];
int dx[8] = {0,0,0,1,1,2,2,2};
int dy[8] = {0,1,2,0,2,0,1,2};
int n, m;
bool in(int x, int y){
    if(x < 1 || x > n) return false;
    if(y < 1 || y > m) return false;
    return true;
}
bool check2(int x, int y){
    for(int i = 0; i < 8; ++i){
        int xx = x + dx[i];
        int yy = y + dy[i];
        if(!in(xx,yy) || s[xx][yy] != '#') return false;
    }
    return true;
}
bool check(int x, int y){
    for(int b = 0; b < 8; ++b){
        if(check2(x-dx[b], y-dy[b]))  return true;
    }
    return false;
}
int main(){

    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i) scanf("%s", s[i]+1);
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j){
            if(s[i][j] == '#'){
                if(!check(i,j)){
                    puts("NO");
                    return 0;
                }
            }
    }
    puts("YES");
    return 0;
}

C：C - Sequence Transformation

题意：给你[1,n]的数，每次会输出整个序列的gcd，然后可以删除一个数，然后问gcd的字典序排列最大是多少。

题解：发现 如果存在奇数，然后我们就会导致 gcd恒为1。 那么我们肯定是先删除奇数。 

然后把奇数删完了之后，剩下了2,4,6,8,我们把这些数都/2，变成了新的1,2,3,4,然后再删除奇数就ok了。递归解决问题。

然后就是 如果出现了 1, 2, 3。 那么肯定是按照顺序删除 1 2 3最好了。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1e5 + 100;
int n;
vector<int> vc;
void dfs(int k, int n){
    if(n == 0) return ;
    if(n == 3){
        printf("%d %d %d", k, k, k*3);
        return ;
    }
    if(n == 1){
        printf("%d ", k);
        return ;
    }
    if(n == 2){
        printf("%d %d", k, 2*k);
        return ;
    }
    for(int i = 1; i <= n; i += 2){
        printf("%d ", k);
    }
    dfs(k*2, n/2);
}
int main(){
    scanf("%d", &n);
    dfs(1, n);
    return 0;
}

D：D - Nature Reserve

题意：有n个保护动物，然后就是让你画一个○，使得所有的动物都在圈内，并且只有1个交点。

不是很想多讲，二分乱搞就好了，注意精度。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1e5 + 100;
pll p[N];
int n, f1 = 0, f2 = 0;
double eps = 1e-9;
bool check2(double x1, double y1, double x2, double y2){
    long double len = y2;
    len *= len;
    long double dis = (long double)(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
    return len >= dis;
}
bool check(double len){
    double lp = -1e7, rp = 1e7;
    for(int i = 1; i <= 100; ++i){
        int lsz = 0, rsz = 0;
        double mid = (lp+rp) / 2;
        for(int k = 1; k <= n; ++k){
            if(check2(p[k].fi,p[k].se,mid,len));
            else if(p[k].fi <= mid ) lsz++;
            else rsz++;
        }
        if(lsz && rsz) return false;
        if(lsz+rsz == 0) return true;
        if(lsz) rp = mid;
        else lp = mid;
    }
    return false;
}
int main(){
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i){
        scanf("%d%d", &p[i].fi, &p[i].se);
        if(p[i].se > 0) ++f1;
        else ++f2, p[i].se = -p[i].se;
    }
    if(f1 && f2){
        puts("-1");
        return 0;
    }
    double l = 0, r = 1e15;
    for(int i = 1; i <= 100; ++i){
        double mid = (l+r)/2;
        if(check(mid)) r = mid;
        else l = mid;
    }
    printf("%.10f", r);
    return 0;
}

E：Split the Tree

题意：给定1棵以1为根的树，让你将这棵数分成竖着的几条链，问链最少是多少条，并且每一条链要求点的个数少于L，路径之和少于S。

题解：

从叶子网上爬，然后通过二分和倍增查看最多能到哪里，然后将这个点指到可以去的那个点为止。

然后不是叶子就接受叶子传上来的最远可以到哪里，如果到不了这个点就把这个点当做新的起点继续去找，如果可以到这个点就记录下最远可以去哪里就好了。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1e5 + 100;
int anc[N][20];
int a[N];
LL sum[N];
int n, l; LL S;
vector<int> vc[N];
int deep[N];
void dfs(int o){
    for(int x : vc[o]){
        deep[x] = deep[o] + 1; sum[x] = sum[o] + a[x];
        anc[x][0] = o;
        for(int i = 1; i < 19; ++i)
            anc[x][i] = anc[anc[x][i-1]][i-1];
        dfs(x);
    }
}
int get(int o, int k){
    for(int i = 19; i >= 0; --i){
        if((k>>i)&1) o = anc[o][i];
    }
    return o;
}
int to[N];
int ans;
int Find(int o){
    int ll = 1, rr = l;
    while(ll <= rr){
        int mid = ll+rr >> 1;
        int p = get(o, mid);
        if(sum[o]-sum[p] <= S) ll = mid+1;
        else rr = mid-1;
    }
    return get(o, ll-1);
}
void solve(int o){
    to[o] = -1;
    for(int x : vc[o]){
        solve(x);
        if(to[x] == -1 || deep[to[x]] >= deep[o]) continue;
        if(to[o] == -1) to[o] = to[x];
        else if(deep[to[o]] > deep[to[x]]) to[o] = to[x];
    }
    if(to[o] == -1) ans++, to[o] = Find(o);
}
int main(){
    scanf("%d%d%lld", &n, &l, &S);
    for(int i = 1; i <= n; ++i){
        scanf("%d", &a[i]);
        if(a[i] > S){
            puts("-1");
            return 0;
        }
    }
    for(int i = 2,v; i <= n; ++i){
        scanf("%d", &v);
        vc[v].pb(i);
    }
    sum[1] = a[1]; deep[1] = 1;
    dfs(1); solve(1);
    cout << ans << endl;
    return 0;
}