Codeforces Round #534 (Div. 2) Solution

Posted 2021-02-05 dup4

 

A. Splitting into digits

Solved.

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int n;
 5 
 6 void solve()
 7 {
 8     printf("%d
", n);
 9     for (int i = 1; i <= n; ++i) printf("%d%c", 1, " 
"[i == n]);
10 }
11 
12 int main()
13 {
14     while (scanf("%d", &n) != EOF)
15         solve();
16     return 0;
17 }

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B. Game with string

Solved.

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define N 100010
 5 char s[N];
 6 
 7 int main()
 8 {
 9     while (scanf("%s", s + 1) != EOF)
10     {
11         stack <char> sta;
12         int cnt = 0;
13         for (int i = 1, len = strlen(s + 1); i <= len; ++i)
14         {
15             if (!sta.empty() && sta.top() == s[i]) sta.pop(), ++cnt;
16             else sta.push(s[i]);
17         }
18         puts(cnt & 1 ? "Yes" : "No");
19     }
20     return 0;
21 }

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C. Grid game

Solved.

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define N 1010
 5 char s[N];
 6 
 7 int main()
 8 {
 9     while (scanf("%s", s + 1) != EOF)
10     {
11         int x = 0, y = 0;
12         for (int i = 1, len = strlen(s + 1); i <= len; ++i)
13         {
14             if (s[i] == ‘1‘)
15             {
16                 printf("%d %d
", 1, x % 2 ? 3 : 1);   
17                 ++x;
18             }
19             else
20             {
21                 printf("%d %d
", 3, y % 4 + 1);
22                 ++y;
23             }
24         }
25     }
26     return 0;
27 }

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D. Game with modulo

Solved.

题意：

交互题

猜数，猜一个$a$

每次询问格式为$(x, y)$

返回结果有两种

$x ;if (x ; mod; a) >= (y ;mod; a)$

$y ;if (x ;mod; a) < (y ;mod; a) $

思路：

我们考虑 从小到大倍增，去找到$a的一个单调范围$

比如说$1, 2, 4, 8, 16, 32  如果某一次询问中返回了x$

那么说明$a在询问的(x, y)范围中  并且2 cdot a 不在这个范围内$

因为是从小到大进行倍增

那么我们考虑某一次询问是$(x, 2cdot x)$

$a在其中$

$如果 a > x, 那么必然有x >= 2 cdot x -  a$

$如果a = x  那么必然有 x ;mod;a = 2 cdot x ;mod; a$

那么这个区间就具有一个单调性，可以进行二分 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 char op[110];
 5 
 6 bool check(int x, int y)
 7 {
 8     printf("? %d %d
", x, y);
 9     fflush(stdout); 
10     scanf("%s", op);
11     return op[0] == ‘y‘;  
12 }
13 
14 void solve(int l, int r)
15 {
16     int base = 1;
17     for (int i = l; i <= r; i += base, base <<= 1)
18     {
19         printf("? %d %d
", i, i + base);
20         fflush(stdout);
21         scanf("%s", op);
22         if (op[0] == ‘x‘)
23         {
24             int l = i + 1, r = i + base - 1, res = -1;
25             while (r - l >= 0)
26             {
27                 int mid = (l + r) >> 1;
28                 if (check(i, mid))
29                 {
30                     l = mid + 1;  
31                     res = mid;
32                 }
33                 else
34                 {
35                     r = mid - 1;
36                 }
37             }    
38             if (res == -1) res = i; 
39             printf("! %d
", res + 1);
40             fflush(stdout);
41             return;
42         }
43     }
44     
45 }
46 
47 int main()
48 {
49     while (scanf("%s", op) && op[0] != ‘e‘)
50         solve(0, 1000000000);    
51     return 0;
52 }

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