标签：+= efi tis pac imp rbegin erase back har
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 Check(int a,int b,int c){ 13 if(a+b>c&&a+c>b&&b+c>a) return 1; 14 if(a+b==c||a+c==b||b+c==a) return 0; 15 return -1; 16 } 17 18 int main(){ 19 #ifndef ONLINE_JUDGE 20 // freopen("1.txt","r",stdin); 21 #endif 22 int a,b,c,d; 23 cin>>a>>b>>c>>d; 24 int flag=-1; 25 flag=max(flag,Check(a,b,c)); 26 flag=max(flag,Check(a,c,d)); 27 flag=max(flag,Check(a,b,d)); 28 flag=max(flag,Check(b,c,d)); 29 if(flag==1) cout<<"TRIANGLE"<<endl; 30 else if(!flag) cout<<"SEGMENT"<<endl; 31 else cout<<"IMPOSSIBLE"<<endl; 32 }
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 char flag; 14 string str[105]; 15 map<char,int>mp; 16 17 int dir[4][2]={0,1,1,0,0,-1,-1,0}; 18 void Check(int x,int y){ 19 for(int i=0;i<4;i++){ 20 int xx=x+dir[i][0]; 21 int yy=y+dir[i][1]; 22 if(xx>=0&&xx<n&&yy>=0&&yy<m){ 23 if(str[xx][yy]!=‘.‘&&str[xx][yy]!=flag) 24 mp[str[xx][yy]]=1; 25 } 26 } 27 } 28 29 int main(){ 30 #ifndef ONLINE_JUDGE 31 freopen("1.txt","r",stdin); 32 #endif 33 cin>>n>>m>>flag; 34 for(int i=0;i<n;i++) cin>>str[i]; 35 for(int i=0;i<n;i++){ 36 for(int j=0;j<m;j++){ 37 if(str[i][j]==flag) Check(i,j); 38 } 39 } 40 cout<<mp.size()<<endl; 41 }
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; 13 14 int a[100005]; 15 16 int main(){ 17 #ifndef ONLINE_JUDGE 18 freopen("1.txt","r",stdin); 19 #endif 20 int ans1=0,ans2=0; 21 cin>>n; 22 for(int i=1;i<=n;i++){ 23 cin>>a[i]; 24 } 25 if(n==1){ 26 cout<<"1 0"<<endl; 27 return 0; 28 } 29 int L=1,R=n; 30 while(L<=R){ 31 if(ans1<=ans2) 32 ans1+=a[L++]; 33 else{ 34 ans2+=a[R--]; 35 } 36 } 37 L--; 38 R++; 39 cout<<L<<" "<<n-R+1<<endl; 40 41 }
因为数据量很小，所以直接暴力搜索（好像也可以用DP）
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,a,b; 13 int h[25]; 14 vector<int>ve,V; 15 int ans=99999999; 16 17 void dfs(int pos,int co){ 18 if(co>=ans) return; 19 if(pos==n){ 20 if(h[pos]<0){ 21 ans=co; 22 V=ve; 23 } 24 return; 25 } 26 for(int i=0;i<=max(h[pos-1]/b+1,max(h[pos]/a+1,h[pos+1]/b+1));i++){ 27 if(h[pos-1]<b*i){ 28 h[pos-1]-=b*i; 29 h[pos+1]-=b*i; 30 h[pos]-=a*i; 31 for(int j=0;j<i;j++) ve.push_back(pos); 32 dfs(pos+1,co+i); 33 for(int j=0;j<i;j++) ve.pop_back(); 34 h[pos-1]+=b*i; 35 h[pos+1]+=b*i; 36 h[pos]+=a*i; 37 } 38 } 39 } 40 41 int main(){ 42 #ifndef ONLINE_JUDGE 43 freopen("1.txt","r",stdin); 44 #endif 45 std::ios::sync_with_stdio(false); 46 cin>>n>>a>>b; 47 for(int i=1;i<=n;i++) cin>>h[i]; 48 dfs(2,0); 49 cout<<ans<<endl; 50 for(int i=0;i<V.size();i++){ 51 cout<<V[i]<<" "; 52 } 53 cout<<endl; 54 }
可以用set或线段树找区间最大最小值，然后不断枚举区间
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 ll n; 13 multiset<ll>se; 14 ll a[100005]; 15 16 int main(){ 17 #ifndef ONLINE_JUDGE 18 freopen("1.txt","r",stdin); 19 #endif 20 ll k; 21 cin>>n>>k; 22 for(int i=1;i<=n;i++){ 23 cin>>a[i]; 24 } 25 ll L=1; 26 ll Max=0; 27 ll tou,wei; 28 for(int i=1;i<=n;i++){ 29 se.insert(a[i]); 30 tou=*se.begin(); 31 wei=*se.rbegin(); 32 while(abs(tou-wei)>k){ 33 se.erase(se.find(a[L++])); 34 tou=*se.begin(); 35 wei=*se.rbegin(); 36 } 37 ll tmp=se.size(); 38 Max=max(Max,tmp); 39 } 40 se.clear(); 41 vector<pair<int,int> >ve; 42 L=1; 43 for(int i=1;i<=n;i++){ 44 se.insert(a[i]); 45 tou=*se.begin(); 46 wei=*se.rbegin(); 47 while(abs(tou-wei)>k){ 48 se.erase(se.find(a[L++])); 49 tou=*se.begin(); 50 wei=*se.rbegin(); 51 } 52 if(se.size()==Max){ 53 ve.push_back(make_pair(L,i)); 54 } 55 } 56 cout<<Max<<" "<<ve.size()<<endl; 57 for(int i=0;i<ve.size();i++){ 58 cout<<ve[i].first<<" "<<ve[i].second<<endl; 59 } 60 }
Codeforces Beta Round #6 (Div. 2 Only)
标签：+= efi tis pac imp rbegin erase back har
原文地址：https://www.cnblogs.com/Fighting-sh/p/10348766.html
