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Codeforces Round #480 (Div. 2)

时间:2018-05-10 00:30:04      阅读:196      评论:0      收藏:0      [点我收藏+]

标签:没有   cout   codeforce   奇数   ble   while   def   .com   color   

A. Links and Pearls

题解:~~~没啥说的,没有pearls也算YES!!!

 1 #pragma warning(disable:4996)
 2 #include<map>
 3 #include<string>
 4 #include<cstdio>
 5 #include<cstring>
 6 #include<iostream>
 7 #include<algorithm>
 8 using namespace std;
 9 #define ll long long
10 
11 const int maxn = 200005;
12 
13 string s;
14 
15 int main()
16 {
17     while (cin >> s) {
18         int n = s.size();
19         int x = 0, y = 0;
20         for (int i = 0; i < n; i++) if (s[i] == o) x++;
21         for (int i = 0; i < n; i++) if (s[i] == -) y++;
22         if (x == 0) printf("YES\n");
23         else {
24             if (y % x == 0) printf("YES\n");
25             else printf("NO\n");
26         }
27         
28     }
29     return 0;
30 }

 B. Marlin

题解:k是偶数,好办。k是奇数,分类讨论,如果k<=n-2,放在中间。如果k>n-2,第三排第1个位置和第n-2个位置放上‘#’(0到n-1)。至于为啥子,拿笔画一下就明白了。

 1 #pragma warning(disable:4996)
 2 #include<map>
 3 #include<string>
 4 #include<cstdio>
 5 #include<cstring>
 6 #include<iostream>
 7 #include<algorithm>
 8 using namespace std;
 9 #define ll long long
10 
11 const int maxn = 200005;
12 
13 int n, k;
14 
15 int main()
16 {
17     while (cin >> n >> k) {
18         if (k % 2 == 0) {
19             printf("YES\n");
20             for (int i = 0; i < n; i++) printf(".");
21             printf("\n");
22             for (int i = 0; i < n; i++) {
23                 if (1 <= i && i <= k / 2) printf("#");
24                 else printf(".");
25             }
26             printf("\n");
27             for (int i = 0; i < n; i++) {
28                 if (1 <= i && i <= k / 2) printf("#");
29                 else printf(".");
30             }
31             printf("\n");
32             for (int i = 0; i < n; i++) printf(".");
33             printf("\n");
34         }
35         else {
36             
37             if (k <= n - 2) {
38                 printf("YES\n");
39                 for (int i = 0; i < n; i++) printf(".");
40                 printf("\n");
41                 for (int i = 0; i < n; i++) {
42                     int pos = n / 2;
43                     int l = pos - k / 2;
44                     int r = l + k - 1;
45                     if (l <= i && i <= r) printf("#");
46                     else printf(".");
47                 }
48                 printf("\n");
49                 for (int i = 0; i < n; i++) printf(".");
50                 printf("\n");
51                 for (int i = 0; i < n; i++) printf(".");
52                 printf("\n");
53             }
54             else {
55                 printf("YES\n");
56                 int t = k - (n - 2);
57                 for (int i = 0; i < n; i++) printf(".");
58                 printf("\n");
59                 for (int i = 0; i < n; i++) {
60                     if (1 <= i && i <= n - 2) printf("#");
61                     else printf(".");
62                 }
63                 printf("\n");
64                 for (int i = 0; i < n; i++) {
65                     if (1 <= i && i <= t - 1) printf("#");
66                     else if (i == n - 2) printf("#");
67                     else printf(".");
68                 }
69                 printf("\n");
70                 for (int i = 0; i < n; i++) printf(".");
71                 printf("\n");    
72                 
73             }
74         }
75     }
76     return 0;
77 }

 C. Posterized

题解:想象一个轴,上面有256个点(0~255),如果当前的点没被染色,则往前找第一个颜色号小于a[i]-k+1的点,然后对这段区间重新染色。

 1 #pragma warning(disable:4996)
 2 #include<map>
 3 #include<string>
 4 #include<cstdio>
 5 #include<cstring>
 6 #include<iostream>
 7 #include<algorithm>
 8 using namespace std;
 9 #define ll long long
10 
11 const int maxn = 100005;
12 
13 int n, k;
14 int color[300];
15 bool use[300];
16 
17 int main()
18 {
19     while (scanf("%d%d", &n, &k) != EOF) {
20         memset(use, 0, sizeof(use));
21         for (int i = 0; i < 256; i++) color[i] = i;
22         while (n--) {
23             int tp;
24             scanf("%d", &tp);
25             int ep = tp;
26             if (!use[tp]) {
27                 while (ep >= 0 && color[ep] >= tp - k + 1) ep--;        //找第一个颜色小于tp-k+1的点
28                 for (int i = ep + 1; i <= tp; i++) {                    //重新染色
29                     use[i] = true;
30                     color[i] = color[ep + 1];
31                 }
32             }
33             cout << color[tp] << " ";
34         }
35     }
36     return 0;
37 }

 

Codeforces Round #480 (Div. 2)

标签:没有   cout   codeforce   奇数   ble   while   def   .com   color   

原文地址:https://www.cnblogs.com/zgglj-com/p/9017381.html

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