Codeforces Round #494 (Div. 3) ABCDE

A 签到题

求出现次数最多的数的次数。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 110;
 5 map<int,int> mp;
 6 int main() {
 7     int n, x, MAX = 0;
 8     cin >> n;
 9     for(int i = 0; i < n; i ++) {
10         cin >> x;
11         mp[x] ++;
12         MAX = max(MAX, mp[x]);
13     }
14     cout << MAX << endl;
15     return 0;
16 }

B. Binary String Constructing

You are given three integers $\mathrm{aa, bb and xx.\; Your\; task\; is\; to\; construct\; a\; binary\; string ss of\; length n=a+bn=a+b such\; that\; there\; are\; exactly aa zeroes,\; exactly bb ones\; and\; exactly xx indices ii (where 1\le i<n1\le i<n)\; such\; that si\ne si+1si\ne si+1.\; It\; is\; guaranteed\; that\; the\; answer\; always\; exists.}$

For example, for the string "01010" there are four indices $\mathrm{ii such\; that 1\le i<n1\le i<n and si\ne si+1si\ne si+1 (i=1,2,3,4i=1,2,3,4).\; For\; the\; string\; "111001"\; there\; are\; two\; such\; indices ii (i=3,5i=3,5).}$

Recall that binary string is a non-empty sequence of characters where each character is either 0 or 1.

Input

The first line of the input contains three integers $\mathrm{aa, bb and xx (1\le a,b\le 100,1\le x<a+b)1\le a,b\le 100,1\le x<a+b).}$

Output

Print only one string $\mathrm{ss,\; where ss is any binary\; string\; satisfying\; conditions\; described\; above.\; It\; is\; guaranteed\; that\; the\; answer\; always\; exists.}$

Examples

input

Copy

2 2 1

output

Copy

1100

input

Copy

3 3 3

output

Copy

101100

input

Copy

5 3 6

output

Copy

01010100

Note

All possible answers for the first example:

• 1100;
• 0011.

All possible answers for the second example:

• 110100;
• 101100;
• 110010;
• 100110;
• 011001;
• 001101;
• 010011;
• 001011.

扫描二维码关注公众号，回复： 1862599 查看本文章

有a个0和b个1，构建a+b个长度的二进制，且有x个位置为x_i != x_i-1

写的有点恶心，写了很多if，构建0101...  或者 101010...

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 110;
 5 char str[N];
 6 int main() {
 7     int a, b, x;
 8     bool flag = 0;
 9     cin >> a >> b >> x;
10     flag = (a>b);
11     for(int i = 1; i <= x; i ++) {
12         if(flag) {
13             if(i&1) {
14                 str[i] = '0';
15                 a--;
16             }
17             else {
18                 str[i] = '1';
19                 b--;
20             }
21         } else {
22             if(i&1) {
23                 str[i] = '1';
24                 b--;
25             }
26             else {
27                 str[i] = '0';
28                 a--;
29             }
30         }
31         
32     }
33     //printf("%s %d %d\n",str+1,a,b);
34     if(flag) {
35         if(x&1) {
36             printf("%s",str+1);
37             for(int i = 0; i < a; i ++) printf("0");
38             for(int i = 0; i < b; i ++) printf("1");    
39         } else {
40             printf("%s",str+1);
41             for(int i = 0; i < b; i ++) printf("1");
42             for(int i = 0; i < a; i ++) printf("0");    
43         }
44     } else {
45         if(x&1) {
46             //101
47             printf("%s",str+1);
48             for(int i = 0; i < b; i ++) printf("1");
49             for(int i = 0; i < a; i ++) printf("0");            
50         } else {
51             //1010
52             printf("%s",str+1);
53             for(int i = 0; i < a; i ++) printf("0");
54             for(int i = 0; i < b; i ++) printf("1");    
55         }
56     }
57     
58     return 0;
59 }

C. Intense Heat

The heat during the last few days has been really intense. Scientists from all over the Berland study how the temperatures and weather change, and they claim that this summer is abnormally hot. But any scientific claim sounds a lot more reasonable if there are some numbers involved, so they have decided to actually calculate some value which would represent how high the temperatures are.

Mathematicians of Berland State University came up with a special heat intensity value. This value is calculated as follows:

Suppose we want to analyze the segment of $\mathrm{nn consecutive\; days.\; We\; have\; measured\; the\; temperatures\; during\; these nn days;\; the\; temperature\; during ii-th\; day\; equals aiai.}$

We denote the average temperature of a segment of some consecutive days as the arithmetic mean of the temperature measures during this segment of days. So, if we want to analyze the average temperature from day $\mathrm{xx to\; day yy,\; we\; calculate\; it\; as y\sum i=xaiy-x+1\sum i=xyaiy-x+1 (note\; that\; division\; is\; performed\; without\; any\; rounding).\; The heat\; intensity\; value is\; the\; maximum\; of average\; temperatures over\; all\; segments\; of\; not\; less\; than kkconsecutive\; days.\; For\; example,\; if\; analyzing\; the\; measures [3,4,1,2][3,4,1,2] and k=3k=3,\; we\; are\; interested\; in\; segments [3,4,1][3,4,1], [4,1,2][4,1,2] and [3,4,1,2][3,4,1,2] (we\; want\; to\; find\; the\; maximum\; value\; of average\; temperature over\; these\; segments).}$

You have been hired by Berland State University to write a program that would compute the heat intensity value of a given period of days. Are you up to this task?

Input

The first line contains two integers $\mathrm{nn and kk (1\le k\le n\le 50001\le k\le n\le 5000)\; —\; the\; number\; of\; days\; in\; the\; given\; period,\; and\; the\; minimum\; number\; of\; days\; in\; a\; segment\; we\; consider\; when\; calculating heat\; intensity\; value,\; respectively.}$

The second line contains $\mathrm{nn integers a1a1, a2a2,\; ..., anan (1\le ai\le 50001\le ai\le 5000)\; —\; the\; temperature\; measures\; during\; given nn days.}$

Output

Print one real number — the heat intensity value, i. e., the maximum of average temperatures over all segments of not less than $\mathrm{kkconsecutive\; days.}$

Your answer will be considered correct if the following condition holds: $|res-res0|<10-6|res-res0|<10-6,\; where resres is\; your\; answer,\; and res0res0 is\; the\; answer\; given\; by\; the\; jury\text{'}s\; solution.$

Example

input

Copy

4 3
3 4 1 2

output

Copy

2.666666666666667

求大于等于k的连续区间的最大平均值。范围很小，直接两个for。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 5010;
 5 int a[N], sum[N];
 6 
 7 int main() {
 8     int n, k, x;
 9     scanf("%d%d",&n,&k);
10     for(int i = 1; i <= n; i ++) {
11         scanf("%d",&a[i]);
12         sum[i] = sum[i-1] + a[i];
13     }
14     double MAX = 0;
15     for(int i = 1; i <= n-k+1; i ++) {
16         for(int j = k+i-1; j <= n; j ++) {
17             int S = sum[j] - sum[i-1];
18             MAX = max(MAX, 1.0*S/(j-i+1));
19         }
20     }
21     printf("%.15lf",MAX);
22     return 0;
23 }

 

D. Coins and Queries

Polycarp has $\mathrm{nn coins,\; the\; value\; of\; the ii-th\; coin\; is aiai.\; It\; is\; guaranteed\; that\; all\; the\; values\; are\; integer\; powers\; of 22 (i.e. ai=2dai=2d for\; some non-negative integer\; number dd).}$

Polycarp wants to know answers on $\mathrm{qq queries.\; The jj-th\; query\; is\; described\; as\; integer\; number bjbj.\; The\; answer\; to\; the\; query\; is\; the\; minimum\; number\; of\; coins\; that\; is\; necessary\; to\; obtain\; the\; value bjbj using\; some\; subset\; of\; coins\; (Polycarp\; can\; use\; only\; coins\; he\; has).\; If\; Polycarp\; can\text{'}t\; obtain\; the\; value bjbj,\; the\; answer\; to\; the jj-th\; query\; is -1.}$

The queries are independent (the answer on the query doesn't affect Polycarp's coins).

Input

The first line of the input contains two integers $\mathrm{nn and qq (1\le n,q\le 2\cdot 1051\le n,q\le 2\cdot 105)\; —\; the\; number\; of\; coins\; and\; the\; number\; of\; queries.}$

The second line of the input contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an —\; values\; of\; coins\; (1\le ai\le 2\cdot 1091\le ai\le 2\cdot 109).\; It\; is\; guaranteed\; that\; all aiai are\; integer\; powers\; of 22 (i.e. ai=2dai=2d for\; some non-negative integer\; number dd).}$

The next $\mathrm{qq lines\; contain\; one\; integer\; each.\; The jj-th\; line\; contains\; one\; integer bjbj —\; the\; value\; of\; the jj-th\; query\; (1\le bj\le 1091\le bj\le 109).}$

Output

Print $\mathrm{qq integers ansjansj.\; The jj-th\; integer\; must\; be\; equal\; to\; the\; answer\; on\; the jj-th\; query.\; If\; Polycarp\; can\text{'}t\; obtain\; the\; value bjbj the\; answer\; to\; the jj-th\; query\; is -1.}$

Example

input

Copy

5 4
2 4 8 2 4
8
5
14
10

output

Copy

1
-1
3
2

有 n 个数，每个数都是2^d，有q次询问，求最少由多少个数组成。

由于每个数都是2^d，所以用数组 a[i] 储存 2ⁱ 的数量，每次询问数字x，就从2³¹开始直到2⁰，只要小于x，就求下最多可以由多少个2ⁱ组成,然后就可以得到答案。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 2e5+10;
 5 int a[40];
 6 map<int,int> mp;
 7 int main() {
 8     int n, q, x;;
 9     scanf("%d%d", &n, &q);
10     for(int i = 1; i <= n; i ++) {
11         scanf("%d", &x);
12         mp[x] ++;
13     }
14     for(int i = 0; i < 32; i ++) {
15         a[i] = mp[1<<i];
16     }
17     while(q--) {
18         scanf("%d",&x);
19         int ans = 0;
20         for(int i = 31; i >= 0; i --) {
21             int tmp = 1<<i;
22             if(tmp <= x) {
23                 int cnt = min(a[i],x/tmp);
24                 ans += cnt;
25                 x -= cnt*tmp;
26             }
27         }
28         if(x == 0) printf("%d\n",ans);
29         else printf("-1\n");
30     }
31     return 0;
32 }

E. Tree Constructing

You are given three integers $\mathrm{nn, dd and kk.}$

Your task is to construct an undirected tree on $\mathrm{nn vertices\; with\; diameter dd and\; degree\; of\; each\; vertex\; at\; most kk,\; or\; say\; that\; it\; is\; impossible.}$

An undirected tree is a connected undirected graph with $\mathrm{n-1n-1 edges.}$

Diameter of a tree is the maximum length of a simple path (a path in which each vertex appears at most once) between all pairs of vertices of this tree.

Degree of a vertex is the number of edges incident to this vertex (i.e. for a vertex $\mathrm{uu it\; is\; the\; number\; of\; edges (u,v)(u,v) that\; belong\; to\; the\; tree,\; where vv is\; any\; other\; vertex\; of\; a\; tree).}$

Input

The first line of the input contains three integers $\mathrm{nn, dd and kk (1\le n,d,k\le 4\cdot 1051\le n,d,k\le 4\cdot 105).}$

Output

If there is no tree satisfying the conditions above, print only one word "NO" (without quotes).

Otherwise in the first line print "YES" (without quotes), and then print $\mathrm{n-1n-1 lines\; describing\; edges\; of\; a\; tree\; satisfying\; the\; conditions\; above.\; Vertices\; of\; the\; tree\; must\; be\; numbered\; from 11 to nn.\; You\; can\; print\; edges\; and\; vertices\; connected\; by\; an\; edge\; in\; any\; order.\; If\; there\; are\; multiple\; answers,\; print\; any\; of\; them.}$

Examples

input

Copy

6 3 3

output

Copy

YES
3 1
4 1
1 2
5 2
2 6

input

Copy

6 2 3

output

Copy

NO

input

Copy

10 4 3

output

Copy

YES
2 9
2 10
10 3
3 1
6 10
8 2
4 3
5 6
6 7

input

Copy

8 5 3

output

Copy

YES
2 5
7 2
3 7
3 1
1 6
8 7
4 3

构建无向图。有n个定点，每个定点最大的度数不能超过k，且直径为d，直径为任意两点直径的距离的最大值。
向构建直径为d的图1-2-3-4-...-d+1,然后从 2 至 d 遍历一遍，第i个顶点可以可以放min(i-1,d+1-i)个，dfs遍历一遍就可以了。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 4e5+10;
 5 vector<int> vs[N];
 6 bool vis[N];
 7 int cnt, n, d, k;
 8 void dfs(int x, int ans) {
 9     if(ans == 0) return;
10     while(vs[x].size() < k && cnt <= n) {
11         vs[x].push_back(cnt);
12         vs[cnt].push_back(x);
13         cnt++;
14         dfs(cnt-1,ans-1);
15     }
16 }
17 int main() {
18     cin >> n >> d >> k;
19     if(d >= n) return 0*printf("NO\n");
20     for(int i = 1; i <= d; i ++) {
21         vs[i].push_back(i+1);
22         vs[i+1].push_back(i);
23     }
24     cnt = d+2;
25     for(int i = 2; i <= d; i ++) {
26         dfs(i,min(i-1,d+1-i));
27     }
28 //    cout << cnt << endl;
29     if(cnt > n) {
30         printf("YES\n");
31         for(int i = 1; i <= n; i ++) {
32             for(int j = 0; j < vs[i].size(); j ++) {
33                 if(!vis[i] || !vis[vs[i][j]]) {
34                     printf("%d %d\n",i,vs[i][j]);
35                     vis[i] = vis[vs[i][j]] = 1;
36                 }
37             }
38         }
39     } else printf("NO\n");
40     return 0;
41 }