Codeforces Round #498 (Div. 3)

这场..

不知道说什么好了唉。。。

A.签到++

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <queue>
 6 #include <map>
 7 #define ll long long
 8 #define out(a) printf("%d ",a)
 9 #define writeln printf("\n")
10 #define N 100050
11 using namespace std;
12 int n;
13 int a[N];
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     n=read();
31     for (int i=1;i<=n;i++)
32       a[i]=read();
33     for (int i=1;i<=n;i++)
34       if (a[i]&1) out(a[i]);
35       else out(a[i]-1);
36     return 0;
37 }

B.n个数,找个方案分成k个区间使每个区间最大值的和最大.

排序找最大的k个数,然后下标减一减就好了。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <queue>
 6 #include <map>
 7 #include <algorithm>
 8 #define ll long long
 9 #define out(a) printf("%d ",a)
10 #define writeln printf("\n")
11 #define N 100050
12 using namespace std;
13 int n,k,tot,ans,now;
14 int b[N];
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 struct dist
30 {
31     int h,id;
32 }a[N];
33 bool cmp(dist a,dist b)
34 {
35     return a.h==b.h?a.id<b.id:a.h>b.h;
36 }
37 int main()
38 {
39     n=read(); k=read();
40     for (int i=1;i<=n;i++)
41       a[i].h=read(),a[i].id=i;
42     sort(a+1,a+n+1,cmp);
43     for (int i=1;i<=k;i++)
44       ans+=a[i].h,b[++tot]=a[i].id;
45       out(ans); writeln;
46       sort(b+1,b+tot+1); now=0;
47     for (int i=1;i<=tot;i++)
48       if (i!=k) out(b[i]-now),now=b[i];
49       else out(n-now);
50     return 0;
51 }

C.n个数分成三个区间,求第一个区间和sum1等于第三个区间和sum3时sum1最大值,区间可以是空集。

预处理出前缀和,后缀和,从两端往中间扫,区间重合的时候break掉.记得开ll

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <queue>
 6 #include <map>
 7 #define ll long long
 8 #define out(a) printf("%lld",a)
 9 #define writeln printf("\n")
10 #define N 200050
11 using namespace std;
12 int n;
13 int a[N];
14 int x,y;
15 ll sum1[N],sum2[N],ans;
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();
33     for (int i=1;i<=n;i++)
34       a[i]=read(),sum1[i]=sum1[i-1]+a[i];
35     for (int i=n;i>=1;i--)
36       sum2[i]=sum2[i+1]+a[i];
37     for (int i=1;i<=n;i++){
38       if (sum1[n]-sum1[i]==sum1[i]){
39         out(sum1[i]);
40         return 0;
41     }
42     }
43     x=1; y=n;
44     while (x<=n&&y>=1){
45         if (x>=y) break;
46       if (sum1[x]==sum2[y]) ans=sum1[x];
47       if (sum1[x]>sum2[y]) y--;
48       else x++;
49     }
50     out(ans);
51     return 0;
52 }

看了下此时排名300多海星,然而接下来就是一个多小时的玩泥巴了。。。

看D,诶这个样例怎么来的....

哦哦哦懂了,好像分类一下就好了,唉好麻烦啊。

诶这个E怎么过这么多了,诶好像xjb写写就能过了.

coding....诶不对啊样例过不了,诶不对啊我想错了GG

滚回去看D,分类好久过样例,无限wa3.。。

唉算了好困回去碎觉了..结果1000多名滚粗,涨分的大好局势毁了..

D,E找时间补补吧。

总结:这场ABC写的海星,手速和思维都够快且失误较少,但是后面的题还是缺少耐心去写导致GG,时间安排也不太好吧..总之div3写不出5题不能算是合格吧.....

原文地址：https://www.cnblogs.com/Kaleidoscope233/p/9326285.html