a题，就不说了吧
b题，直接从大到小排序1-limit的所有数的lowbit，再从大到小贪心组成sum就行了
1 #include<cstdio>
2 #include<iostream>
3 #include<algorithm>
4 #include<cstring>
5 #define N 200000
6 using namespace std;
7 int pos[N],a[N],s[N],f[N],la[N],b[N],i,j,k,ans,n,p,sum,lim;
8 bool fl[N];
9 struct node{
10 int x,y;
11 }u[N];
12 int pow(int x, int n)
13 {
14 int result;
15 if (n == 0)
16 return 1;
17 else
18 {
19 while ((n & 1) == 0)
20 {
21 n >>= 1;
22 x *= x;
23 }
24 }
25 result = x;
26 n >>= 1;
27 while (n != 0)
28 {
29 x *= x;
30 if ((n & 1) != 0)
31 result *= x;
32 n >>= 1;
33 }
34 return result;
35 }
36 bool cmp(node a, node b)
37 {
38 return a.y>b.y;
39 }
40 int main()
41 {
42 scanf("%d%d",&sum,&lim);
43 for (i=1; i<=lim; i++)
44 {
45 j=i;
46 k=0;
47 while (j % 2==0)
48 {
49 ++k;
50 j=j / 2;
51 }
52 pos[k]=i;
53 u[i].x=i;
54 u[i].y=pow(2,k);
55 }
56 sort(u+1,u+lim+1,cmp);
57
58 i=1;
59 while (sum)
60 {
61 while (sum < u[i].y) i++;
62 sum-=u[i].y;
63 a[++ans]=u[i].x;
64 i++;
65 if (i > lim && sum)
66 {
67 printf("-1");
68 return 0;
69 }
70 }
71 printf("%d\n",ans);
72 for (i=1; i<=ans; i++)
73 printf("%d ",a[i]);
74 return 0;
75 }
c题，直接找每连接两端中值较小的部分
1 #include<cstdio>
2 #include<iostream>
3 #include<algorithm>
4 using namespace std;
5 int n,m,ans,i,x,y,a[10000];
6 int main()
7 {
8 scanf("%d%d",&n,&m);
9 for (i=1; i<=n; i++)
10 scanf("%d",&a[i]);
11 for (i=1; i<=m; i++)
12 {
13 scanf("%d%d",&x,&y);
14 ans+=min(a[x], a[y]);
15 }
16 printf("%d",ans);
17 return 0;
18 }
d题，并查集处理
1 #include<cstdio>
2 #include<iostream>
3 #include<algorithm>
4 #define N 200000
5 using namespace std;
6 typedef long long LL;
7 struct edge{
8 LL u,v,c;
9 }e[N];
10 int n,m;
11 LL f[N],s[N],x[N],i,fa,fb;
12 double sum;
13 int find(LL x){
14 if (x != f[x]) f[x]=find(f[x]);
15 return f[x];
16 }
17 bool cmp(edge a,edge b){
18 return a.c>b.c;
19 }
20 int main()
21 {
22 scanf("%d%d",&n,&m);
23 for (i=1; i<=n; i++){
24 scanf("%I64d",&x[i]);
25 f[i]=i, s[i]=1;
26 }
27 for (i=1; i<=m; i++){
28 scanf("%I64d%I64d",&e[i].u,&e[i].v);
29 e[i].c=min(x[e[i].u], x[e[i].v]);
30 }
31 sort(e+1,e+m+1,cmp);
32 for (i=1; i<=m; i++){
33 fa=find(e[i].u), fb=find(e[i].v);
34 if (fa==fb) continue;
35 sum+=(s[fa] * s[fb]) * e[i].c;
36 f[fb]=fa, s[fa]+=s[fb];
37 }
38 sum=2.0 * (sum / n) / (n-1);
39 printf("%.6lf",sum);
40 return 0;
41 }
codeforces Round #250 (div2),布布扣,bubuko.com
时间： 2024-12-23 02:56:15