Codeforces Beta Round #72 (Div. 2 Only)
http://codeforces.com/contest/84
A
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define lson l,mid,rt<<1
4 #define rson mid+1,r,rt<<1|1
5 #define sqr(x) ((x)*(x))
6 #define pb push_back
7 #define eb emplace_back
8 #define maxn 1000006
9 #define eps 1e-8
10 #define pi acos(-1.0)
11 #define rep(k,i,j) for(int k=i;k<j;k++)
12 typedef long long ll;
13 typedef unsigned long long ull;
14 int main(){
15 #ifndef ONLINE_JUDGE
16 // freopen("input.txt","r",stdin);
17 #endif
18 std::ios::sync_with_stdio(false);
19 ll n;
20 cin>>n;
21 cout<<2*n-n/2<<endl;
22 }
B
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define lson l,mid,rt<<1
4 #define rson mid+1,r,rt<<1|1
5 #define sqr(x) ((x)*(x))
6 #define pb push_back
7 #define eb emplace_back
8 #define maxn 1000006
9 #define eps 1e-8
10 #define pi acos(-1.0)
11 #define rep(k,i,j) for(int k=i;k<j;k++)
12 typedef long long ll;
13 typedef unsigned long long ull;
14
15 int n;
16 ll a[100005];
17
18 int main(){
19 #ifndef ONLINE_JUDGE
20 // freopen("input.txt","r",stdin);
21 #endif
22 std::ios::sync_with_stdio(false);
23 cin>>n;
24 for(int i=1;i<=n;i++) cin>>a[i];
25 ll ans=n;
26 ll co=1;
27 for(int i=1;i<n;i++){
28 if(a[i]==a[i+1]){
29 co++;
30 }
31 else{
32 ans+=co*(co-1)/2;
33 co=1;
34 }
35 }
36 ans+=co*(co-1)/2;
37 cout<<ans<<endl;
38 }
C
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define lson l,mid,rt<<1
4 #define rson mid+1,r,rt<<1|1
5 #define sqr(x) ((x)*(x))
6 #define pb push_back
7 #define eb emplace_back
8 #define maxn 1000006
9 #define eps 1e-8
10 #define pi acos(-1.0)
11 #define rep(k,i,j) for(int k=i;k<j;k++)
12 typedef long long ll;
13 typedef unsigned long long ull;
14
15 const int N = 10010, D = 20000;
16 int n, m, x, y, c[N], r[N], s[N];
17 vector<int> v[D * 2 + 10];
18
19 int main(){
20 #ifndef ONLINE_JUDGE
21 // freopen("input.txt","r",stdin);
22 #endif
23 std::ios::sync_with_stdio(false);
24 cin>>n;
25 int ans=0;
26 for (int i = 1; i <= n; i++){
27 cin>>c[i]>>r[i];
28 c[i] += D; s[i] = -1;
29 for (int j = c[i] - r[i]; j <= c[i] + r[i]; j++) v[j].push_back(i);
30 }
31 cin>>m;
32 for (int i = 1; i <= m; i++){
33 cin>>x>>y;
34 x += D;
35 for (auto j : v[x]) if (s[j] == -1)
36 if (y * y + (x - c[j]) * (x - c[j]) <= r[j] * r[j]){
37 s[j] = i;
38 ans++;
39 }
40 }
41 cout<<ans<<endl;
42 for (int i = 1; i <= n; i++) cout<<s[i]<<" ";
43 }
D
二分+模拟
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define lson l,mid,rt<<1
4 #define rson mid+1,r,rt<<1|1
5 #define sqr(x) ((x)*(x))
6 #define pb push_back
7 #define eb emplace_back
8 #define maxn 1000006
9 #define eps 1e-8
10 #define pi acos(-1.0)
11 #define rep(k,i,j) for(int k=i;k<j;k++)
12 typedef long long ll;
13 typedef unsigned long long ull;
14
15 ll a[100005];
16 ll n,k;
17 int book[100005];
18 vector<int>ans;
19
20 bool erfen(ll mid){
21 ll sum=0;
22 for(int i=1;i<=n;i++){
23 sum+=min(a[i],mid);
24 }
25 if(sum>=k) return true;
26 return false;
27 }
28
29 int main(){
30 #ifndef ONLINE_JUDGE
31 // freopen("input.txt","r",stdin);
32 #endif
33 std::ios::sync_with_stdio(false);
34 cin>>n>>k;
35 ll sum=0;
36 ll Max=0;
37 ll L,R,mid;
38 for(int i=1;i<=n;i++){
39 cin>>a[i];
40 sum+=a[i];
41 Max=max(Max,a[i]);
42 }
43 if(sum<k) cout<<-1<<endl;
44 else if(sum==k){}
45 else{
46 L=0,R=Max,mid;
47 while(L<=R){
48 mid=L+R>>1;
49 if(erfen(mid)){
50 R=mid-1;
51 }
52 else{
53 L=mid+1;
54 }
55 }
56 ll kk=R;
57 sum=0;
58 for(int i=1;i<=n;i++){
59 sum+=min(a[i],kk);
60 a[i]-=min(a[i],kk);
61 }
62 int flag=1;
63 for(int i=1;i<=n;i++){
64 if(sum>k&&a[i]>0){
65 book[i]=1;
66 ans.pb(i);
67 }
68 else if(a[i]>0){
69 sum++;
70 a[i]--;
71 if(sum>k&&flag&&a[i]>=0){
72 flag=0;
73 ans.pb(i);
74 book[i]=1;
75 }
76 }
77
78 }
79 for(int i=1;i<=n;i++){
80 if(!book[i]&&a[i]>0){
81 book[i]=1;
82 ans.pb(i);
83 }
84 }
85 for(int i=0;i<ans.size();i++){
86 cout<<ans[i]<<" ";
87 }
88 }
89
90 }
E
__builtin_popcount() 函数，计算一个数转二进制之后有几个1
一道很好的最短路的题目。
题意：求S到T的最短路径，其中，路径上不同的字母不能超过K个，且答案的字典序要最小，不存在输出-1
思路：因为K小于等于4,所以可以用状压来存放不同字母的个数。在优先队列中存放距离，路径的字符串，K和该点的下表，跑一次最短路即可。
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define lson l,mid,rt<<1
4 #define rson mid+1,r,rt<<1|1
5 #define sqr(x) ((x)*(x))
6 #define pb push_back
7 #define eb emplace_back
8 #define maxn 1000006
9 #define eps 1e-8
10 #define pi acos(-1.0)
11 #define rep(k,i,j) for(int k=i;k<j;k++)
12 typedef long long ll;
13 typedef pair<int,int> pii;
14 typedef pair<pair<int,string>,pii> ppp;
15 typedef unsigned long long ull;
16
17 int n,m,k;
18 int S,T;
19 string s[55];
20 set<pii>se;
21
22 int dist(int a,int b){
23 return abs(a/m-b/m)+abs(a%m-b%m);
24 }
25
26 int dir[4][2]={0,1,1,0,0,-1,-1,0};
27
28 void Dijstra(){
29 priority_queue<ppp,vector<ppp>,greater<ppp>>Q;
30 Q.push({{dist(S,T),""},{0,S}});
31 ppp ss;
32 string str;
33 int book,pos;
34 int x,y,xx,yy;
35 while(!Q.empty()){
36 ss=Q.top();
37 Q.pop();
38 str=ss.first.second;
39 book=ss.second.first;
40 pos=ss.second.second;
41 if(!se.count({book,pos})){
42 se.insert({book,pos});
43 x=pos/m,y=pos%m;
44 for(int i=0;i<4;i++){
45 xx=x+dir[i][0];
46 yy=y+dir[i][1];
47 if(xx>=0&&xx<n&&yy>=00&&yy<m){
48 if(s[xx][yy]==‘T‘){
49 cout<<str<<endl;
50 exit(0);
51 }
52 if(__builtin_popcount(book|(1<<(s[xx][yy]-‘a‘)))<=k)
53 Q.push({{dist(xx*m+yy,T)+str.length()+1,str+s[xx][yy]},{book|(1<<(s[xx][yy]-‘a‘)),xx*m+yy}});
54 }
55 }
56 }
57 }
58
59 }
60
61 int main(){
62 #ifndef ONLINE_JUDGE
63 // freopen("input.txt","r",stdin);
64 #endif
65 std::ios::sync_with_stdio(false);
66 cin>>n>>m>>k;
67 for(int i=0;i<n;i++){
68 cin>>s[i];
69 for(int j=0;j<m;j++){
70 if(s[i][j]==‘S‘){
71 S=i*m+j;
72 }
73 if(s[i][j]==‘T‘){
74 T=i*m+j;
75 }
76 }
77 }
78 Dijstra();
79 cout<<-1<<endl;
80 }
原文地址：https://www.cnblogs.com/Fighting-sh/p/10489781.html
时间： 2024-10-02 06:27:53