Codeforces Round #494 (Div. 3)爆炸记

Posted 2020-12-02 kaleidoscope233

一开始网站爆炸,502刷新了10min,那时就知道这场要gg了。。。
A.签到题
B.求一个01串包含a个0,b个1且恰好有x个位置s[i]不等于s[i+1];
考虑a,b大小关系确定第一个是0还是1,前面构造x-1个10或01,后面就连着输出相同的就好了。
没看到样例wa3两次。。。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <queue>
 6 #include <cmath>
 7 #define ll long long
 8 #define out(a) printf("%d",a)
 9 #define writeln printf("
")
10 #define min(a,b) a<b?a:b;
11 #define max(a,b) a>b?a:b;
12 using namespace std;
13 int n,m,a,b,c,d,x,ans,now;
14 int read()
15 {
16     int s=0,t=1; char c;
17     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
18     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
19     return s*t;
20 }
21 ll readl()
22 {
23     ll s=0,t=1; char c;
24     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
25     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
26     return s*t;
27 }
28 int main()
29 {
30     b=read(); a=read(); x=read(); 
31     if (x==1) {
32       for (int i=1;i<=a;i++)
33         out(1);
34       for (int i=1;i<=b;i++)
35         out(0);
36       return 0;
37     }
38     if (a>b) {
39     for (int i=1;i<=x;i++)
40       if (i&1) out(1),c++,now=1;
41       else out(0),d++,now=0;
42     }
43     else {
44     for (int i=1;i<=x;i++)
45       if (i&1) out(0),d++,now=0;
46       else out(1),c++,now=1;
47     }
48       a-=c; b-=d;
49       if (now==1) {
50         for (int i=1;i<=a;i++)
51           out(1);
52         for (int i=1;i<=b;i++)
53           out(0);
54       }
55       else {
56         for (int i=1;i<=b;i++)
57           out(0);
58          for (int i=1;i<=a;i++)
59           out(1);
60       }
61     return 0;
62 }

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C.找一个不小于k的区间平均值最大.
读错题意罚时爆炸QAQ。
记录前缀和然后二重循环扫一遍即可。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <queue>
 6 #include <cmath>
 7 #define ll long long
 8 #define out(a) printf("%lld",a)
 9 #define writeln printf("
")
10 #define min(a,b) a<b?a:b;
11 #define max(a,b) a>b?a:b;
12 using namespace std;
13 int n,m,a,b,x,k,num;
14 double ans;
15 int sum[100050];
16 int read()
17 {
18     int s=0,t=1; char c;
19     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
20     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
21     return s*t;
22 }
23 ll readl()
24 {
25     ll s=0,t=1; char c;
26     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
27     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
28     return s*t;
29 }
30 int main()
31 {
32     n=read(); k=read(); 
33     for (int i=1;i<=n;i++) {
34       x=read(); sum[i]=sum[i-1]+x;
35     }
36     for (int j=k;j<=n;j++) {
37         num=-2333333;
38     for (int i=j;i<=n;i++) 
39       num=max(num,sum[i]-sum[i-j]);
40     if ((double)num/j>ans) ans=(double)num/j;
41     }
42     printf("%.15f",ans);
43     return 0;
44 }

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D.给n个2的d次方的数,m个询问问是否有n个数的子集和等于x.
正解都想出来了,可就是没写出来,细节总是错错错QAQ
考虑把所有数都搞成2进制,如果这一位不够的话就往下一位拆。。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <queue>
 6 #include <cmath>
 7 #define ll long long
 8 #define out(a) printf("%d",a)
 9 #define writeln printf("
")
10 using namespace std;
11 int n,k;
12 int x,cnt,m,tot,ans;
13 int a[200050],b[200050];
14 bool flag;
15 int read()
16 {
17     int s=0,t=1; char c;
18     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
19     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
20     return s*t;
21 }
22 ll readl()
23 {
24     ll s=0,t=1; char c;
25     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
26     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
27     return s*t;
28 }
29 int main()
30 {
31     n=read(),k=read();
32     for (int i=1;i<=n;i++){
33       x=read(); cnt=0;
34       while (true) {
35         if (x==1||x==0) break;
36         cnt++; x/=2;
37       }
38       a[cnt]++; //out(cnt); writeln;
39     }
40     for (int i=1;i<=k;i++){
41       m=read(); flag=false; tot=ans=0; b[0]=0;
42       while (true) {
43         b[++tot]=m%2;
44         if (flag) break;
45         m/=2;
46         if (m==0||m==1) flag=true;
47       }
48       for (int j=tot;j>=0;j--)
49         if (a[j-1]>=b[j]) ans+=b[j],b[j]=0;
50         else {
51           b[j]-=a[j-1];
52           b[j-1]+=b[j]*2;
53           ans+=a[j-1];
54         }
55       if (b[0]>0) out(-1);
56       else out(ans); 
57       writeln;
58     }
59     return 0;
60 }

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总结
1.码力还是太差了,细节问题总出错。
2.心态问题,急忙交题不测样例是大忌,被一些bug卡的很烦。
3.读题能力差,现在由于cf复制的问题不太兹磁翻译,所以基本上都是读英语或dalao翻译,题意总是看错。
There is still a long way to go....