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AtCoder Beginner Contest 349

A

直接累加输出负数

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//by Totoro
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define frep(i,a,n) for(int i=a;i>=n;i--)
#define int long long
#define PII pair<int,int>
#define lowbit(x) (x&(-x))
const int mod=1e9+7;
const double pai=acos(-1.0);
#define ios ios::sync_with_stdio(0); cin.tie(0),cout.tie(0);
#define LF(x)   fixed<<setprecision(x)
const int N=1e6+100;
#define CY cout<<"YES<<'\n"
#define CN cout<<"NO"<<'\n'
int a[N];
void Totoro()
{
int n;
cin>>n;
int sum=0;
rep(i,1,n)
{
    cin>>a[i];
    sum+=a[i];
}
cout<<-sum<<'\n';
}
signed main()
{
    int t=1;
    while(t--)
    {
        Totoro();
    }
    return 0;
}

B

两个map，先记录次数，再记录次数出现的字母，在判断

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//by Totoro
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define frep(i,a,n) for(int i=a;i>=n;i--)
#define int long long
#define PII pair<int,int>
#define lowbit(x) (x&(-x))
const int mod=1e9+7;
const double pai=acos(-1.0);
#define ios ios::sync_with_stdio(0); cin.tie(0),cout.tie(0);
#define LF(x)   fixed<<setprecision(x)
const int N=1e6+100;
#define CY cout<<"YES<<'\n"
#define CN cout<<"NO"<<'\n'
int a[N];
void Totoro()
{
string s;
cin>>s;
int n=s.length();
s=" "+s; 
map<char,int>a;
rep(i,1,n)
{
a[s[i]]++;
}
map<int,char>b;
for(auto &[x,y]:a)
{
    b[y]++;
}
for(auto &[x,y]:b)
{
    if(y!=2&&y!=0)
    {
        cout<<"No";
        return ;
    }
}
cout<<"Yes";

}
signed main()
{
    int t=1;
    while(t--)
    {
        Totoro();
    }
    return 0;
}

C

本题有点抽象，因为没看到后缀爆wa，需要好好擦眼睛了。

只需要对有x和没有x进行讨论，主要就是看有没有出现，记录一下即可。

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//by Totoro
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define frep(i,a,n) for(int i=a;i>=n;i--)
#define int long long
#define PII pair<int,int>
#define lowbit(x) (x&(-x))
const int mod=1e9+7;
const double pai=acos(-1.0);
#define ios ios::sync_with_stdio(0); cin.tie(0),cout.tie(0);
#define LF(x)   fixed<<setprecision(x)
const int N=1e6+100;
#define CY cout<<"YES<<'\n"
#define CN cout<<"NO"<<'\n'
void Totoro()
{
string s;
string x;
cin>>s>>x;

int n=s.length()-1;
int b=x.length()-1;

    x[0]+='a'-'A';
 x[1]+='a'-'A';
  x[2]+='a'-'A';
if(x[2]=='x')
{
    map<char,int>mp;
    int c=-1,d=-1;
    bool flag=1;
    bool f=1;
    rep(i,0,n)
{
if(s[i]==x[0]&&flag)
{
    c=i;
    flag=0;
}
if(s[i]==x[1])
{
d=i;
}
}
if(c!=-1&&d!=-1&&c<d)
{
    cout<<"Yes";
}
else
{
    cout<<"No";
}
}
else
{
    int c=-1,d=-1;
    bool flag=1;
    bool f=0;
    rep(i,0,n)
{
if(s[i]==x[0]&&flag)
{
    c=i;
    flag=0;
}
if(s[i]==x[2])
{
d=i;
}
}
if(c!=-1&&d!=-1)
{
rep(i,c+1,d-1)
{
    if(s[i]==x[1])
    {
        f=1;
    }
}
}
    if(f)
    {
        cout<<"Yes";
    }
    else
    {
        cout<<"No";
    }
}


}
signed main()
{
    int t=1;
    while(t--)
    {
        Totoro();
    }
    return 0;
}

D

这个条件决定了取模等于0这一特判，从高到低位的贪心的特判从此而来

\(S(2^i j, 2^i (j+1))\)

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//by Totoro
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define frep(i,a,n) for(int i=a;i>=n;i--)
#define int long long
#define PII pair<int,int>
#define lowbit(x) (x&(-x))
const int mod=1e9+7;
const double pai=acos(-1.0);
#define ios ios::sync_with_stdio(0); cin.tie(0),cout.tie(0);
#define LF(x)   fixed<<setprecision(x)
const int N=1e6+100;
#define CY cout<<"YES<<'\n"
#define CN cout<<"NO"<<'\n'
int a[N];
void Totoro()
{
int l,r;
cin>>l>>r;
vector<ll>ans;
while(l<r)
{
    ll nowl=0;
    for(int i=60;i>=0;i--)
{
        //贪心
    if(l%(1ll<<i)==0ll&&l+(1ll<<i)<=r)
        //一个取模特判，一个边界特判
{
nowl=(l+(1ll<<i));
break;
}
}
ans.push_back(l);
ans.push_back(nowl);
l=nowl;
}
cout<<ans.size()/2<<'\n';
for(int i=0;i<ans.size();i+=2)
{
    cout<<ans[i]<<" "<<ans[i+1]<<'\n';
}
}
signed main()
{
    int t=1;
    while(t--)
    {
        Totoro();
    }
    return 0;
}