Tink off Challenge Final Round 2017

A：热手题

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <string>
 6 #include <cstring>
 7 #include <cmath>
 8 #include <map>
 9 #include <stack>
10 #include <set>
11 #include <vector>
12 #include <queue>
13 #include <time.h>
14 #define eps 1e-7
15 #define INF 0x3f3f3f3f
16 #define MOD 1000000007
17 #define rep0(j,n) for(int j=0;j<n;++j)
18 #define rep1(j,n) for(int j=1;j<=n;++j)
19 #define pb push_back
20 #define set0(n) memset(n,0,sizeof(n))
21 #define ll long long
22 #define ull unsigned long long
23 #define iter(i,v) for(edge *i=head[v];i;i=i->nxt)
24 #define max(a,b) (a>b?a:b)
25 #define min(a,b) (a<b?a:b)
26 #define print_runtime printf("Running time:%.3lfs\n",double(clock())/1000.0)
27 #define TO(j) printf(#j": %d\n",j);
28 //#define OJ
29 using namespace std;
30 const int MAXINT = 100010;
31 const int MAXNODE = 100010;
32 const int MAXEDGE = 2*MAXNODE;
33 char BUF,*buf;
34 int read(){
35     char c=getchar();int f=1,x=0;
36     while(!isdigit(c)){if(c=='-') f=-1;c=getchar();}
37     while(isdigit(c)){x=x*10+c-'0';c=getchar();}
38     return f*x;
39 }
40 char get_ch(){
41     char c=getchar();
42     while(!isalpha(c)) c=getchar();
43     return c;
44 }
45 //------------------- Head Files ----------------------//
46 
47 int a,b,c,n,t,ans;
48 void get_input();
49 void work();
50 int main() {
51     get_input();
52     work();
53     return 0;
54 }
55 void work(){
56     printf("%d\n",ans);
57 }
58 void get_input(){
59     a=read();b=read();c=read();
60     n=read();
61     rep0(i,n){
62         t=read();
63         if(t>b&&t<c) ans++; 
64     }
65 }
A：少女祈祷中

 

B：初中数学

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <string>
 6 #include <cstring>
 7 #include <cmath>
 8 #include <map>
 9 #include <stack>
10 #include <set>
11 #include <vector>
12 #include <queue>
13 #include <time.h>
14 #define eps 1e-7
15 #define INF 0x3f3f3f3f
16 #define MOD 1000000007
17 #define rep0(j,n) for(int j=0;j<n;++j)
18 #define rep1(j,n) for(int j=1;j<=n;++j)
19 #define pb push_back
20 #define set0(n) memset(n,0,sizeof(n))
21 #define ll long long
22 #define ull unsigned long long
23 #define iter(i,v) for(edge *i=head[v];i;i=i->nxt)
24 #define max(a,b) (a>b?a:b)
25 #define min(a,b) (a<b?a:b)
26 #define print_runtime printf("Running time:%.3lfs\n",double(clock())/1000.0)
27 #define TO(j) printf(#j": %d\n",j);
28 //#define OJ
29 using namespace std;
30 const int MAXINT = 100010;
31 const int MAXNODE = 100010;
32 const int MAXEDGE = 2*MAXNODE;
33 char BUF,*buf;
34 int read(){
35     char c=getchar();int f=1,x=0;
36     while(!isdigit(c)){if(c=='-') f=-1;c=getchar();}
37     while(isdigit(c)){x=x*10+c-'0';c=getchar();}
38     return f*x;
39 }
40 char get_ch(){
41     char c=getchar();
42     while(!isalpha(c)) c=getchar();
43     return c;
44 }
45 //------------------- Head Files ----------------------//
46 
47 int n,h;
48 void get_input();
49 void work();
50 int main() {
51     get_input();
52     work();
53     return 0;
54 }
55 void work(){
56     double s=0.5*h,ans = s/n;
57     for(int i=1;i<n;i++){
58         printf("%.12lf\n",h*sqrt(i*ans/s));
59     }
60 }
61 void get_input(){
62     n=read();h=read();
63 }
B：少女祈祷中

 

C：我讨厌游戏题

首先双方都只会选对他来说最优的那些字母来放置

为了方便，我们把希望字典序大的一方称为A，另一方为B

当A的最大字符比B的最小字符要大的时候，双方都会把对自己最优往前放

而当A的最大字符比B的最小字符要小的时候，双方会尽量把对自己不优的字符往后放，逼迫对方把不优的字符往前放

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <string>
 6 #include <cstring>
 7 #include <cmath>
 8 #include <map>
 9 #include <stack>
10 #include <set>
11 #include <vector>
12 #include <queue>
13 #include <time.h>
14 #define eps 1e-7
15 #define INF 0x3f3f3f3f
16 #define MOD 1000000007
17 #define rep0(j,n) for(int j=0;j<n;++j)
18 #define rep1(j,n) for(int j=1;j<=n;++j)
19 #define pb push_back
20 #define set0(n) memset(n,0,sizeof(n))
21 #define ll long long
22 #define ull unsigned long long
23 #define iter(i,v) for(edge *i=head[v];i;i=i->nxt)
24 #define max(a,b) (a>b?a:b)
25 #define min(a,b) (a<b?a:b)
26 #define print_runtime printf("Running time:%.3lfs\n",double(clock())/1000.0)
27 #define TO(j) printf(#j": %d\n",j);
28 //#define OJ
29 using namespace std;
30 const int MAXINT = 300010;
31 const int MAXNODE = 100010;
32 const int MAXEDGE = 2*MAXNODE;
33 char BUF,*buf;
34 int read(){
35     char c=getchar();int f=1,x=0;
36     while(!isdigit(c)){if(c=='-') f=-1;c=getchar();}
37     while(isdigit(c)){x=x*10+c-'0';c=getchar();}
38     return f*x;
39 }
40 char get_ch(){
41     char c=getchar();
42     while(!isalpha(c)) c=getchar();
43     return c;
44 }
45 //------------------- Head Files ----------------------//
46 
47 char s[2][MAXINT],ans[MAXINT];
48 int cnt[2][26],p1,p2,c1,c2,l;
49 void get_input();
50 void work();
51 int main() {
52     get_input();
53     work();
54     return 0;
55 }
56 void work(){
57     p1=0;p2=0;
58     rep0(i,l){
59         if(s[1][p2]<=s[0][p1]){
60             for(int j=l-1;j>=i;j--){
61                 if(((l-1-j)^i)&1){
62                     ans[j]=s[1][--c2];
63                 }else{
64                     ans[j]=s[0][--c1];
65                 }
66             }
67             break;
68         }
69         if(i&1){//big
70             ans[i]=s[1][p2++];
71         }else{
72             ans[i]=s[0][p1++];
73         }
74     }
75     ans[l]='\0';
76     printf("%s\n",ans);
77 }
78 void get_input(){
79     rep0(i,2){
80         scanf("%s",s[i]);
81         l = strlen(s[i]);
82     }
83     c1 = l/2+l%2;
84     c2 = l/2;
85     sort(s[0],s[0]+l);
86     sort(s[1],s[1]+l,greater<char>() );
87 }
C：少女祈祷中

D：这大概是最难的一题。。。

首先把每个点本身加入自己的临接表里，然后我们声称所有的有着相同临接表的点的权值都相同，显然这样不会使情况更坏

而且显然临接表不同的点权值一定不同

然后我们把权值相同的点缩成一个点，重新构图，我们发现如果一个点的度数大于2那肯定不存在一个合法的解，而且如果这些点成环了也不存在合法的解，最后只剩下一条链的情况，直接一路标下来即可

注意哈希冲突，没有stl这道题不是要写死

  1 #include <iostream>
  2 #include <cstdlib>
  3 #include <cstdio>
  4 #include <algorithm>
  5 #include <string>
  6 #include <cstring>
  7 #include <cmath>
  8 #include <map>
  9 #include <stack>
 10 #include <set>
 11 #include <vector>
 12 #include <queue>
 13 #include <time.h>
 14 #define eps 1e-7
 15 #define INF 0x3f3f3f3f
 16 #define MOD 1000000007
 17 #define rep0(j,n) for(int j=0;j<n;++j)
 18 #define rep1(j,n) for(int j=1;j<=n;++j)
 19 #define pb push_back
 20 #define set0(n) memset(n,0,sizeof(n))
 21 #define ll long long
 22 #define ull unsigned long long
 23 #define iter(i,v) for(edge *i=head[v];i;i=i->nxt)
 24 #define max(a,b) (a>b?a:b)
 25 #define min(a,b) (a<b?a:b)
 26 #define print_runtime printf("Running time:%.3lfs\n",double(clock())/1000.0)
 27 #define TO(j) printf(#j": %d\n",j);
 28 //#define OJ
 29 using namespace std;
 30 const int MAXINT = 100010;
 31 const int MAXNODE = 300010;
 32 const int MAXEDGE = 2 * MAXNODE;
 33 char BUF, *buf;
 34 int read() {
 35     char c = getchar(); int f = 1, x = 0;
 36     while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
 37     while (isdigit(c)) {x = x * 10 + c - '0'; c = getchar();}
 38     return f * x;
 39 }
 40 char get_ch() {
 41     char c = getchar();
 42     while (!isalpha(c)) c = getchar();
 43     return c;
 44 }
 45 //------------------- Head Files ----------------------//
 46 int cnt, n, m,cnt_e;
 47 vector<int> e[MAXNODE], ne[MAXNODE];
 48 map<pair<int,int>, vector<int> > t;
 49 map<pair<int,int>, int> ys;
 50 set<pair<int,int> > es;
 51 int d[MAXNODE],ans[MAXNODE],cl[MAXNODE];
 52 pair<int,int> hs[MAXNODE];
 53 void dfs(int p,int fa){
 54     for(auto i = ne[p].begin();i!=ne[p].end();i++){
 55         if(*i==fa) continue;
 56         cl[*i]=cl[p]+1;
 57         dfs(*i,p);
 58     }
 59 }
 60 void get_input();
 61 void work();
 62 int main() {
 63     get_input();
 64     work();
 65     return 0;
 66 }
 67 void work() {
 68     for (auto i = t.begin(); i != t.end(); i++) {
 69         ys[i->first] = ++cnt;
 70     }
 71     rep1(i, n) {
 72         for (auto j = e[i].begin(); j != e[i].end(); j++) {
 73             int u = ys[hs[i]],v=ys[hs[*j]];
 74             if(u==v) continue;
 75             if(u>v) swap(u,v);
 76             if(es.count(make_pair(u,v))) continue;
 77             ne[u].push_back(v);
 78             ne[v].push_back(u);
 79             d[u]++;
 80             d[v]++;
 81             if(d[u]>2||d[v]>2) {
 82                 printf("NO\n");
 83                 return;
 84             }
 85             cnt_e++;
 86             es.insert(make_pair(u,v));
 87         }
 88     }
 89     if(cnt_e==cnt) {
 90         printf("NO\n");
 91         return;
 92     }
 93     rep1(i,cnt) cl[i]=1;
 94     rep1(i,cnt) if(d[i]==1) {dfs(i,0); break;}
 95     for (auto i = t.begin(); i != t.end(); i++) {
 96         int x = cl[ys[i->first]];
 97         for(auto j = i->second.begin();j!=i->second.end();j++){
 98             ans[*j]=x;
 99         }
100     }
101     printf("YES\n");
102     rep1(i,n) printf("%d ",ans[i]);
103     putchar('\n');
104 }
105 void get_input() {
106     n = read(); m = read();
107     rep0(i, m) {
108         int u = read(), v = read();
109         e[u].push_back(v);
110         e[v].push_back(u);
111     }
112     rep1(i, n) {
113         e[i].push_back(i);
114         sort(e[i].begin(), e[i].end());
115         int v = 0,v2 = 0;
116         for (auto j = e[i].begin(); j != e[i].end(); j++) {
117             v = ((ll)v * 19260817ll + *j) % MOD;
118             v2 = ((ll)v2 * 171ll + *j) % MOD;
119         }
120         t[make_pair(v,v2)].push_back(i);
121         hs[i] = make_pair(v,v2);
122     }
123 }
D：少女祈祷中

E：有一个很有趣的性质

如果剩下的胡萝卜个数是偶数，那么最终的答案是中间两个的较大值

如果剩下的胡萝卜个数是奇数，那么如果最中央胡萝卜比它两侧的胡萝卜都要小，最终答案是中央胡萝卜，否则是中间三个的中间数

有了这个直接维护两个堆就可以了，注意最后一个胡萝卜需要特判

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <string>
 6 #include <cstring>
 7 #include <cmath>
 8 #include <map>
 9 #include <stack>
10 #include <set>
11 #include <vector>
12 #include <queue>
13 #include <time.h>
14 #define eps 1e-7
15 #define INF 0x3f3f3f3f
16 #define MOD 1000000007
17 #define rep0(j,n) for(int j=0;j<n;++j)
18 #define rep1(j,n) for(int j=1;j<=n;++j)
19 #define pb push_back
20 #define set0(n) memset(n,0,sizeof(n))
21 #define ll long long
22 #define ull unsigned long long
23 #define iter(i,v) for(edge *i=head[v];i;i=i->nxt)
24 #define print_runtime printf("Running time:%.3lfs\n",double(clock())/1000.0)
25 #define TO(j) printf(#j": %d\n",j);
26 //#define OJ
27 using namespace std;
28 const int MAXINT = 300010;
29 const int MAXNODE = 100010;
30 const int MAXEDGE = 2*MAXNODE;
31 char BUF,*buf;
32 int read(){
33     char c=getchar();int f=1,x=0;
34     while(!isdigit(c)){if(c=='-') f=-1;c=getchar();}
35     while(isdigit(c)){x=x*10+c-'0';c=getchar();}
36     return f*x;
37 }
38 char get_ch(){
39     char c=getchar();
40     while(!isalpha(c)) c=getchar();
41     return c;
42 }
43 //------------------- Head Files ----------------------//
44 int n,a[MAXINT],m[MAXINT];
45 priority_queue<int> even,odd;
46 void get_input();
47 void work();
48 int main() {
49     get_input();
50     work();
51     return 0;
52 }
53 void work(){
54     if(n==1){
55         printf("%d\n",a[1]);
56         return ;
57     }
58     if(n&1) even.push(a[n/2+1]);
59     rep0(i,n){
60         if((n-i)&1){
61             if(i==n-1){
62                 int ans=0;
63                 rep1(i,n) ans=max(ans,a[i]);
64                 printf("%d ",ans);
65                 continue;
66             }
67             if(n&1){
68                 odd.push(m[n/2+1+i/2]);
69                 odd.push(m[n/2+1-i/2]);
70             }else{
71                 odd.push(m[n/2+1+i/2]);
72                 odd.push(m[n/2-i/2]);
73             }
74             printf("%d ",odd.top());
75         }else {
76             if(n&1){
77                 even.push(a[n/2+1-(i+1)/2]);
78                 even.push(a[n/2+1+(i+1)/2]);
79             }else{
80                 even.push(a[n/2-i/2]);
81                 even.push(a[n/2+1+i/2]);
82             }
83             printf("%d ",even.top());
84         }
85     }
86     putchar('\n');
87 }
88 void get_input(){
89     n=read();
90     rep1(i,n) a[i]=read();
91     m[1]=a[1];m[n]=a[n];
92       for(int i=2;i<=n-1;i++){
93           if(a[i]>a[i+1]&&a[i]>a[i-1]) m[i] = max(a[i-1],a[i+1]);
94           else m[i]=a[i];
95       }
96 }
E：少女祈祷中

F：这题真是没话说，接近O(60*nlogn)是怎么过去的

发现这个可以很容易的用线段树维护，只需要维护每个数字在这里面的倍数即可（在十位数上出现即为10，百位即为100）

然后打tag的时候暴力O(10)修改

我还傻乎乎的加了一堆乱七八糟的东西，尝试了一下利用区间覆盖会导致的合并元素段，结果反而T了

  1 #include <iostream>
  2 #include <cstdlib>
  3 #include <cstdio>
  4 #include <algorithm>
  5 #include <string>
  6 #include <cstring>
  7 #include <cmath>
  8 #include <map>
  9 #include <stack>
 10 #include <set>
 11 #include <vector>
 12 #include <queue>
 13 #include <time.h>
 14 #define eps 1e-7
 15 #define INF 0x3f3f3f3f
 16 #define MOD 1000000007
 17 #define rep0(j,n) for(int j=0;j<n;++j)
 18 #define rep1(j,n) for(int j=1;j<=n;++j)
 19 #define pb push_back
 20 #define set0(n) memset(n,0,sizeof(n))
 21 #define ll long long
 22 #define ull unsigned long long
 23 #define iter(i,v) for(edge *i=head[v];i;i=i->nxt)
 24 #define print_runtime printf("Running time:%.3lfs\n",double(clock())/1000.0)
 25 #define TO(j) printf(#j": %d\n",j);
 26 //#define OJ
 27 using namespace std;
 28 const int MAXINT = 100010;
 29 const int MAXNODE = 100010;
 30 const int MAXEDGE = 2 * MAXNODE;
 31 char BUF, *buf;
 32 int read() {
 33     char c = getchar(); int f = 1, x = 0;
 34     while (!isdigit(c)) { if (c == '-') f = -1; c = getchar(); }
 35     while (isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }
 36     return f * x;
 37 }
 38 char get_ch() {
 39     char c = getchar();
 40     while (!isalpha(c)) c = getchar();
 41     return c;
 42 }
 43 //------------------- Head Files ----------------------//
 44 int cnt, a[MAXINT], n, q;
 45 struct node {
 46     node *l, *r;
 47     int lb, rb, tag[10], mid, cnt;
 48     ll mul[10], sum;
 49     void pushup() {
 50         rep0(i, 10) mul[i] = l->mul[i] + r->mul[i];
 51         sum = l->sum + r->sum;
 52     }
 53     void addtag(int from, int to) {
 54         tag[from] = to;
 55         pushdown();
 56     }
 57     void pushdown() {
 58         ll ns[10] = {};
 59         if (rb - lb >1) {
 60             rep0(i, 10) l->tag[i] = tag[l->tag[i]];
 61             rep0(i, 10) r->tag[i] = tag[r->tag[i]];
 62         }
 63         rep0(i, 10) ns[tag[i]] += mul[i];
 64         rep0(i, 10) mul[i] = ns[i];
 65         sum = 0;
 66         rep0(i, 10) sum += i * mul[i];
 67         rep0(i, 10) tag[i] = i;
 68     }
 69 };
 70 struct Segtree {
 71     node mp[MAXINT * 4];
 72     int cnt;
 73     node *root;
 74     Segtree() {
 75         cnt = 0;
 76     }
 77     node *newnode(int lb, int rb) {
 78         node *p = &mp[cnt++];
 79         p->lb = lb;
 80         p->rb = rb;
 81         p->mid = (lb + rb) / 2;
 82         p->sum = 0;
 83         rep0(i, 10) p->tag[i] = i;
 84         rep0(i, 10) p->mul[i] = 0;
 85         return p;
 86     }
 87     void build(int l, int r) {
 88         build(l, r, root);
 89     }
 90     void build(int l, int r, node *&p) {
 91         p = newnode(l, r);
 92         if (r - l > 1) {
 93             build(l, p->mid, p->l);
 94             build(p->mid, r, p->r);
 95             p->pushup();
 96         }
 97         else {
 98             int t = 1;
 99             while (a[l]) {
100                 p->mul[a[l] % 10] += t;
101                 a[l] /= 10;
102                 t = t * 10;
103             }
104             rep0(i, 10) p->sum += i * p->mul[i];
105         }
106     }
107     void modify(int l, int r, int from, int to) {
108         modify(l, r, from, to, root);
109     }
110     void modify(int l, int r, int from, int to, node *p) {
111         p->pushdown();
112         if (p->lb >= r || p->rb <= l) return;
113         if (p->lb >= l && p->rb <= r) {
114             p->addtag(from, to);
115             return;
116         }
117         modify(l, r, from, to, p->l);
118         modify(l, r, from, to, p->r);
119         p->pushup();
120     }
121     ll query(int l, int r) {
122         return query(l, r, root);
123     }
124     ll query(int l, int r, node *p) {
125         p->pushdown();
126         if (p->lb >= r || p->rb <= l) return 0;
127         if (p->lb >= l && p->rb <= r) return p->sum;
128         return query(l, r, p->l) + query(l, r, p->r);
129     }
130 } seg;
131 void get_input();
132 void work();
133 int main() {
134     get_input();
135     work();
136     return 0;
137 }
138 void work() {
139     while (q--) {
140         int op = read();
141         if (op == 1) {
142             int l = read(), r = read(), from = read(), to = read();
143             if (from == to) continue;
144             seg.modify(l, r + 1, from, to);
145         }
146         else {
147             int l = read(), r = read();
148             printf("%lld\n", seg.query(l, r + 1));
149         }
150     }
151 }
152 void get_input() {
153     n = read(); q = read();
154     rep1(i, n) a[i] = read();
155     seg.build(1, n + 1);
156 }
157 /*
158 5 5
159 38 43 4 12 70
160 1 1 3 4 8
161 2 2 4
162 1 4 5 0 8
163 1 2 5 8 7
164 2 1 5
165 */
F：少女祈祷中

 

posted @ 2017-06-10 17:44  LoveYayoi  阅读(190)  评论(0)  编辑  收藏  举报