A：

题意：问可以休息多少次。

代码：

#include
      
       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
       
         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：

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
      
       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
       
         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：

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
      
       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
       
         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
        
          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：

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
      
       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
       
         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：

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
      
       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
       
         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
        
          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;}