Codeforces Beta Round #12 (Div 2 Only)
http://codeforces.com/contest/12
A
水题
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 maxn 1000010
7 typedef long long ll;
8 /*#ifndef ONLINE_JUDGE
9 freopen("1.txt","r",stdin);
10 #endif */
11
12 string str[5];
13
14 int main(){
15 #ifndef ONLINE_JUDGE
16 freopen("1.txt","r",stdin);
17 #endif
18 for(int i=0;i<3;i++){
19 cin>>str[i];
20 }
21 for(int i=0;i<3;i++){
22 for(int j=0;j<3;j++){
23 if(str[i][j]!=str[2-i][2-j]){
24 cout<<"NO"<<endl;
25 return 0;
26 }
27 }
28 }
29 cout<<"YES"<<endl;
30
31
32 return 0;
33 }
B
水题
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 maxn 1000010
7 typedef long long ll;
8 /*#ifndef ONLINE_JUDGE
9 freopen("1.txt","r",stdin);
10 #endif */
11
12 string str[5];
13 int ch[15];
14
15 int main(){
16 #ifndef ONLINE_JUDGE
17 // freopen("1.txt","r",stdin);
18 #endif
19 string str1,str2,ans="";
20 cin>>str1>>str2;
21 for(int i=0;i<str1.length();i++){
22 ch[str1[i]-‘0‘]++;
23 }
24 for(int i=1;i<10;i++){
25 if(ch[i]){
26 ans+=char(i+‘0‘);
27 ch[i]--;
28 break;
29 }
30 }
31
32 for(int i=0;i<10;i++){
33 for(int j=0;j<ch[i];j++){
34 ans+=char(i+‘0‘);
35 }
36 }
37 // cout<<ans<<" "<<str2<<endl;
38 if(ans==str2) cout<<"OK"<<endl;
39 else cout<<"WRONG_ANSWER"<<endl;
40 return 0;
41 }
C
水题
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 maxn 1000010
7 typedef long long ll;
8 /*#ifndef ONLINE_JUDGE
9 freopen("1.txt","r",stdin);
10 #endif */
11
12 int n,m;
13 map<string,int>mp;
14
15 int a[105];
16 bool cmp1(int a,int b){
17 return a>b;
18 }
19
20 bool cmp2(int a,int b){
21 return a<b;
22 }
23
24 bool cmp(pair<int,string>a,pair<int,string>b){
25 return a.first>b.first;
26 }
27
28 vector<pair<int,string> >ve;
29
30 int main(){
31 #ifndef ONLINE_JUDGE
32 freopen("1.txt","r",stdin);
33 #endif
34 cin>>n>>m;
35 string str;
36 for(int i=0;i<n;i++) cin>>a[i];
37 for(int i=0;i<m;i++){
38 cin>>str;
39 mp[str]++;
40 }
41 map<string,int>::iterator it;
42 for(it=mp.begin();it!=mp.end();it++){
43 ve.push_back(make_pair(it->second,it->first));
44 }
45 sort(a,a+n,cmp2);
46 int ans1=0,ans2=0;
47 sort(ve.begin(),ve.end(),cmp);
48 /* for(int i=0;i<ve.size();i++){
49 cout<<ve[i].first<<" "<<ve[i].second<<endl;
50 }*/
51 for(int i=0;i<ve.size();i++){
52 ans1+=ve[i].first*a[i];
53 }
54 sort(a,a+n,cmp1);
55 for(int i=0;i<ve.size();i++){
56 ans2+=ve[i].first*a[i];
57 }
58 cout<<ans1<<" "<<ans2<<endl;
59
60 }
D
线段树好题
题意：n个女性比三种属性，一旦有一个女的三种属性都被另一个女的压制，那这个女的会自杀，问多少女性会自杀
思路：
先按第一个属性从大到小严格排序，然后再把第二个属性离散化作为线段树的下标，最后再把第三个属性作为线段树上的值，然后查询最大值即可
如果第一个属性有相等的情况，需要把相等的情况全部比较完再更新
1 #include <cstdio>
2 #include <iostream>
3 #include <cstring>
4 #include <algorithm>
5 #define lson num<<1,s,mid
6 #define rson num<<1|1,mid+1,e
7 #define maxn 500005
8
9 using namespace std;
10
11 struct node
12 {
13 int a,b,c;
14 bool operator < (const node &cmp)const
15 {
16 if(a!=cmp.a)return a<cmp.a;
17 if(b!=cmp.b)return b<cmp.b;
18 return c<cmp.c;
19 }
20 }wm[maxn];
21
22 int res[maxn<<2];
23 int x[maxn];
24
25 void pushup(int num)
26 {
27 res[num]=max(res[num<<1],res[num<<1|1]);
28 }
29 void build(int num,int s,int e)
30 {
31 res[num]=-1;
32 if(s==e)return ;
33 int mid=(s+e)>>1;
34 build(lson);build(rson);
35 }
36 void update(int num,int s,int e,int pos,int val)
37 {
38 if(s==e)
39 {
40 res[num]=max(res[num],val);
41 return;
42 }
43 int mid=(s+e)>>1;
44 if(pos<=mid)update(lson,pos,val);
45 else update(rson,pos,val);
46 pushup(num);
47 }
48 int query(int num,int s,int e,int l,int r)
49 {
50 if(l<=s && r>=e)return res[num];
51 int mid=(s+e)>>1;
52 if(r<=mid)return query(lson,l,r);
53 else if(l>mid)return query(rson,l,r);
54 else return max(query(lson,l,mid),query(rson,mid+1,r));
55 }
56 int main()
57 {
58 int n;
59 scanf("%d",&n);
60 for(int i=1;i<=n;i++)
61 scanf("%d",&wm[i].a);
62 for(int i=1;i<=n;i++)
63 {
64 scanf("%d",&wm[i].b);
65 x[i]=wm[i].b;
66 }
67 for(int i=1;i<=n;i++)
68 scanf("%d",&wm[i].c);
69
70 sort(wm+1,wm+1+n);
71
72 sort(x+1,x+1+n);
73
74 int m=unique(x+1,x+1+n)-x;
75 m--;
76
77 int last=wm[n].a;
78 int r=n;
79 int l=n;
80 int ans=0;
81
82 wm[0].a=0x3f3f3f3f;
83
84 for(int i=n;i>=1;)
85 {
86 while(wm[l].a==last)
87 {
88 l--;
89 }
90 int c=r;
91 while(c>l)
92 {
93 int pos=lower_bound(x+1,x+m+1,wm[c].b)-x;
94 if(pos+1<=m && query(1,1,m,pos+1,m)>wm[c].c)ans++;
95 c--;
96 }
97 c=r;
98 while(c>l)
99 {
100 int pos=lower_bound(x+1,x+m+1,wm[c].b)-x;
101 update(1,1,m,pos,wm[c].c);
102 c--;
103 }
104 i=l;r=l;last=wm[i].a;
105 }
106 printf("%d\n",ans);
107 return 0;
108 }
E
构造题
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 maxn 1000010
7 typedef long long ll;
8 /*#ifndef ONLINE_JUDGE
9 freopen("1.txt","r",stdin);
10 #endif */
11
12 int book[1005][1005];
13
14 int main(){
15 #ifndef ONLINE_JUDGE
16 // freopen("1.txt","r",stdin);
17 #endif
18 int n;
19 std::ios::sync_with_stdio(false);
20 cin>>n;
21 for(int i = 0 ; i < n-1 ; i++){
22 for(int j = 0 ; j < n-1 ; j++)
23 book[i][j] = (i+j)%(n-1)+1;
24 }
25 for(int i = 0 ; i < n ; i++){
26 book[i][n-1] = book[i][i];
27 book[n-1][i] = book[i][i];
28 book[i][i] = 0;
29 }
30 for(int i = 0 ; i < n ; i++){
31 for(int j = 0 ; j < n ; j++)
32 cout<<book[i][j]<<" ";
33 cout<<endl;
34 }
35
36 }
原文地址：https://www.cnblogs.com/Fighting-sh/p/10352997.html
时间： 2024-10-27 08:01:33