Codeforces Round #849 (Div. 4) ABCDEFG1
https://codeforces.com/contest/1791
ABCDEG1全水题，直接上代码
F往下翻
A. Codeforces Checking
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=31;
const LL N=1e6+10,M=4002;
const double PI=3.1415926535;
#define endl '\n'
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
int T=1;
cin>>T;
while(T--)
{
char c;
cin>>c;
string s="codeforces";
bool flag=false;
for(int i=0;i<s.size();i++)
{
if(s[i]==c)
{
flag=true;
break;
}
}
if(flag==false) cout<<"NO"<<endl;
else cout<<"YES"<<endl;
}
return 0;
}
B. Following Directions
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=31;
const LL N=1e6+10,M=4002;
const double PI=3.1415926535;
#define endl '\n'
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
int T=1;
cin>>T;
while(T--)
{
LL n;
cin>>n;
string s;
cin>>s;
LL x=0,y=0;
bool flag=false;
for(int i=0;i<s.size();i++)
{
if(s[i]=='U') y++;
else if(s[i]=='D') y--;
else if(s[i]=='L') x--;
else x++;
if(x==1&&y==1) flag=true;
}
if(flag==true) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}
C .Prepend and Append
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=31;
const LL N=1e6+10,M=4002;
const double PI=3.1415926535;
#define endl '\n'
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
int T=1;
cin>>T;
while(T--)
{
LL n;
cin>>n;
string s;
cin>>s;
LL len=s.size();
LL sum=len;
for(int i=0,j=len-1;i<j;i++,j--)
{
if(s[i]=='0'&&s[j]=='1') sum-=2;
else if(s[i]=='1'&&s[j]=='0') sum-=2;
else break;
}
cout<<sum<<endl;
}
return 0;
}
D. Distinct Split
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=31;
const LL N=1e6+10,M=4002;
const double PI=3.1415926535;
#define endl '\n'
LL a[N]={0},b[N]={0};
int main()
{
//cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
int T=1;
cin>>T;
while(T--)
{
LL n;
cin>>n;
string s;
cin>>s;
s=" "+s;
LL len=s.size();
map<char,LL> mp;
a[0]=b[len]=0;
for(int i=1;i<s.size();i++)
{
mp[s[i]]++;
a[i]=a[i-1]+(mp[s[i]]==1?1:0);
}
mp.clear();
for(int i=len-1;i>=1;i--)
{
mp[s[i]]++;
b[i]=b[i+1]+(mp[s[i]]==1?1:0);
}
LL maxn=0;
for(int i=1;i<s.size()-1;i++)
{
//cout<<a[i]<<" "<<b[i]<<endl;
maxn=max(maxn,a[i]+b[i+1]);
}
cout<<maxn<<endl;
mp.clear();
}
return 0;
}
E. Negatives and Positives
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=31;
const LL N=1e6+10,M=4002;
const double PI=3.1415926535;
#define endl '\n'
LL a[N]={0},b[N]={0};
int main()
{
//cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
int T=1;
cin>>T;
while(T--)
{
LL n;
cin>>n;
LL fu=0,zero=0,zh=0;
for(int i=1;i<=n;i++)
{
cin>>a[i];
if(a[i]<0) fu++;
else if(a[i]==0) zero++;
else zh++;
}
LL sum=0;
if(fu%2==0||zero!=0)
{
for(int i=1;i<=n;i++)
sum+=abs(a[i]);
cout<<sum<<endl;
}
else
{
LL minn=abs(a[1]);
for(int i=1;i<=n;i++)
{
minn=min(abs(a[i]),minn);
sum+=abs(a[i]);
}
//cout<<sum<<" "<<minn<<endl;
cout<<sum-2*minn<<endl;
}
}
return 0;
}
G1. Teleporters (Easy Version)
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=31;
const LL N=1e6+10,M=4002;
const double PI=3.1415926535;
#define endl '\n'
LL a[N]={0},b[N]={0};
int main()
{
//cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
int T=1;
cin>>T;
while(T--)
{
LL n,m;
cin>>n>>m;
for(int i=1;i<=n;i++)
{
cin>>a[i];
b[i]=i+a[i];
}
sort(b+1,b+1+n);
LL sum=0;
for(int i=1;i<=n;i++)
{
if(m>=b[i]) m-=b[i],sum++;
else break;
}
cout<<sum<<endl;
}
return 0;
}
补题:
谁能想到我的F竟然被hack了，傻狗爆哭，还是少写点究极暴力[囧]
F. Range Update Point Query
https://codeforces.com/contest/1791/problem/F
题目大意：
给定一个数组a，q个更新，2种查询状态:
1 l r ————l≤i≤r中的所有数字，将ai值更新为ai每个位置上的数的总和。
2 x ————输出a[x]。
input
3
5 8
1 420 69 1434 2023
1 2 3
2 2
2 3
2 4
1 2 5
2 1
2 3
2 5
2 3
9999 1000
1 1 2
2 1
2 2
1 1
1
2 1
output
6
15
1434
1
6
7
36
1
1
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=31;
const LL N=1e6+10,M=4010;
const double PI=3.1415926535;
#define endl '\n'
LL solve(LL x)//位数求和
{
LL sum=0;
while(x)
{
sum+=x%10;
x/=10;
}
return sum;
}
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
int T=1;
cin>>T;
while(T--)
{
LL n,q;
cin>>n>>q;
vector<LL> a(n+1);
set<LL> st;
for(int i=1;i<=n;i++)
{
cin>>a[i];
if(a[i]>=10) st.insert(i);//大于等于10，一旦被指定，则需要更改，标记位置
}
while(q--)
{
LL op;
cin>>op;
LL x,y;
if(op==1)
{
cin>>x>>y;
while(!st.empty())
{
auto it=st.lower_bound(x);
if(*it>y||it==st.end()) break;
x=(*it)+1;
a[*it]=solve(a[*it]);
if(a[*it]<=9) st.erase(it);
}
}
else
{
cin>>x;
cout<<a[x]<<endl;
}
}
}
return 0;
}
