Codeforces Round 640 (Div. 4) ABCDEFG
https://codeforces.com/contest/1352
不知道怎么的复制过来的代码容易歪，观看效果可能不大好。
这场古早div4，大题极其友好，除了E卡空间卡到我爆炸，别的都体验感极好。
A. Sum of Round Numbers
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN,INF=0x3f3f3f3f;
const LL N=2e7+10,M=2023;
const LL mod=100000007;
const double PI=3.1415926535;
#define endl '\n'
LL a[6]={0,10,100,1000,10000};
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
cin>>T;
while(T--)
{
LL n;
cin>>n;
vector<LL> v;
for(int i=4;i>=1;i--)
{
if(n>a[i])
{
v.push_back(n/a[i]*a[i]);
n-=n/a[i]*a[i];
}
}
if(n!=0)
{
v.push_back(n);
n=0;
}
cout<<v.size()<<endl;
for(int i=0;i<v.size();i++)
{
cout<<v[i]<<" ";
}
cout<<endl;
v.clear();
}
return 0;
}
B. Same Parity Summands
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN,INF=0x3f3f3f3f;
const LL N=2e7+10,M=2023;
const LL mod=100000007;
const double PI=3.1415926535;
#define endl '\n'
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
cin>>T;
while(T--)
{
LL n,m;
cin>>n>>m;
if(n<m) cout<<"NO"<<endl;
else if(n%m==0)
{
cout<<"YES"<<endl;
for(int i=1;i<=m;i++)
{
cout<<n/m<<" ";
}
cout<<endl;
}
else
{
if((n-(m-1))%2==1&&n-(m-1)>0)//1 1 1
{
cout<<"YES"<<endl;
for(int i=1;i<=m;i++)
{
if(i!=m) cout<<1<<" ";
else cout<<n-(m-1)<<endl;
}
}
else if((n-(m-1)*2)%2==0&&n-(m-1)*2>0)//2 2 2
{
cout<<"YES"<<endl;
for(int i=1;i<=m;i++)
{
if(i!=m) cout<<2<<" ";
else cout<<(n-(m-1)*2)<<endl;
}
}
else cout<<"NO"<<endl;
}
}
return 0;
}
C. K-th Not Divisible by n
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN,INF=0x3f3f3f3f;
const LL N=2e7+10,M=2023;
const LL mod=100000007;
const double PI=3.1415926535;
#define endl '\n'
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
cin>>T;
while(T--)
{
LL n,k;
cin>>n>>k;
if(k<n) cout<<k<<endl;
else if(k==n) cout<<n+1<<endl;
else
{
LL last=k/n;
LL sum1=k;
LL sum2=k+last;
LL ans=k+last;
//cout<<last<<" "<<sum1<<" "<<sum2<<" "<<ans<<endl;
while(last!=0)
{
last=sum2/n-sum1/n;
ans+=last;
sum1=sum2;
sum2=ans;
//cout<<last<<" "<<sum1<<" "<<sum2<<" "<<ans<<endl;
}
cout<<ans<<endl;
}
}
return 0;
}
D. Alice, Bob and Candies
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN,INF=0x3f3f3f3f;
const LL N=2e7+10,M=2023;
const LL mod=100000007;
const double PI=3.1415926535;
#define endl '\n'
LL a[N];
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
cin>>T;
while(T--)
{
LL n;
cin>>n;
LL res=0;
for(int i=1;i<=n;i++)
{
cin>>a[i];
res+=a[i];
}
LL alice=0,bob=0;
LL alast=0,blast=0;
LL t=0;
for(int i=1,j=n; ; )
{
alast=0;
while(i<=j&&alast<=blast)
{
alast+=a[i++];
}
alice+=alast;
t++;
if(i>j) break;
blast=0;
while(i<=j&&blast<=alast)
{
blast+=a[j--];
}
bob+=blast;
t++;
if(i>j) break;
}
cout<<t<<" "<<alice<<" "<<res-alice<<endl;
}
return 0;
}
E. Special Elements
题目大意：
给定一个长度为n的数组a，问我们能不能选取某一段的数字之和，然后同时数组中也存在这个和？
这样的和有几个？
input
5
9
3 1 4 1 5 9 2 6 5
3
1 1 2
5
1 1 1 1 1
8
8 7 6 5 4 3 2 1
1
1
output
5
1
0
4
0
memory limit per test：64 megabytes
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
//const LL MAXN=1e18,MINN=-MAXN,INF=0x3f3f3f3f;
const LL N=1e5+10,M=2023;
//const LL mod=100000007;
//const double PI=3.1415926535;
//#define endl '\n'
int main()
{
//cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
cin>>T;
while(T--)
{
int n;
//cin>>n;
scanf("%d",&n);
LL mp[N]={0},b[N];
for(int i=1;i<=n;i++)
{
//cin>>a[i];
scanf("%d",&b[i]);
mp[b[i]]++;
b[i]+=b[i-1];
}
int sum=0;
for(int i=1;i<=n;i++)
{
for(int j=i+1;j<=n;j++)
{
int res=b[j]-b[i-1];
if(res<=n&&mp[res])//给定元素个数<=8000且元素总和<=8000的数组
{
sum+=mp[res];
mp[res]=0;//只能加一次
}
}
}
//cout<<sum<<endl;
printf("%d\n",sum);
}
return 0;
}
F. Binary String Reconstruction
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN,INF=0x3f3f3f3f;
const LL N=1e6+10,M=2023;
const LL mod=100000007;
const double PI=3.1415926535;
#define endl '\n'
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
cin>>T;
while(T--)
{
LL a,b,c;
cin>>a>>b>>c;
if(b==0)
{
if(a!=0)
{
for(int i=1;i<=a+1;i++)
{
cout<<"0";
}
}
else if(c!=0)
{
for(int i=1;i<=c+1;i++)
{
cout<<"1";
}
}
cout<<endl;
}
else
{
string ans="";
if(b!=0)
{
for(int i=1;i<=b;i++)
{
if(i==1) ans+="10";
else
{
if(i%2==0) ans+="1";
else ans+="0";
}
}
}
string sum="";
for(int i=0;i<ans.size();i++)
{
if(ans[i]=='1'&&c!=0)
{
for(int i=1;i<=c;i++)
{
sum+="1";
}
c=0;
}
else if(ans[i]=='0'&&a!=0)
{
for(int i=1;i<=a;i++)
{
sum+="0";
}
a=0;
}
sum+=ans[i];
}
cout<<sum<<endl;
}
}
return 0;
}
G. Special Permutation
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN,INF=0x3f3f3f3f;
const LL N=1e6+10,M=2023;
const LL mod=100000007;
const double PI=3.1415926535;
#define endl '\n'
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
cin>>T;
while(T--)
{
LL n;
cin>>n;
if(n==2||n==3) cout<<"-1"<<endl;
else if(n==4) cout<<"3 1 4 2"<<endl;
else if(n==5) cout<<"5 3 1 4 2"<<endl;
else if(n==6) cout<<"5 3 6 2 4 1"<<endl;
else if(n==7) cout<<"5 1 3 6 2 4 7"<<endl;
else
{
deque<LL> dq;
dq.push_back(5); dq.push_back(1); dq.push_back(3);
dq.push_back(6); dq.push_back(2); dq.push_back(4);
dq.push_back(7);
for(int i=8;i<=n;i++)
{
if(i%2==0) dq.push_front(i);
else dq.push_back(i);
}
for(int i=0;i<dq.size();i++)
{
cout<<dq[i]<<" ";
}
cout<<endl;
dq.clear();
}
}
return 0;
}
