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 }