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AtCoder Beginner Contest 346

A - Adjacent Product

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#include<bits/stdc++.h>
using namespace std;

int main(){
  int n;cin>>n;
  vector<int> v(n);
  for(int i=0;i<n;i++){
    cin>>v[i];
  }
  for(int i=1;i<n;i++) cout<<v[i]*v[i-1]<<' ';
  cout<<'\n';
}

B - Piano

数据好小

暴力出奇迹

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#include <bits/stdc++.h>
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define frep(i, a, n) for (int i = a; i >= n; i--)
using ll = long long;
using namespace std;
const int N=1e6+100;
int sum1[N];
int sum2[N];
int main()
{
    string s="wbwbwwbwbwbw";
    string a=" ";
    int x,y;
    cin>>x>>y;
    rep(i,1,100)
    {
        a+=s;
    }
    int t=s.length()-1;
    rep(i,1,t*100)
{
if(a[i]=='w')
{
    sum1[i]=sum1[i-1]+1;
    sum2[i]=sum2[i-1];
}
else
{
    sum1[i]=sum1[i-1];
    sum2[i]=sum2[i-1]+1;
}
}
rep(i,x+y,t*100-1)
{
if(sum1[i]-sum1[i-x-y]==x&&sum2[i]-sum2[i-x-y]==y)
{cout<<"Yes";return 0;
}
}
cout<<"No";

}

C - Σ

留下那些小于的排序去重

减去即可，都是一眼

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#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=2e5+5;
int n,k,a[N],R,ans;
signed main(){
	cin>>n>>k;
	for(int i=1,x;i<=n;i++){
		cin>>x;
		if(x>k) continue;
		a[++R]=x;
	}
	sort(a+1,a+R+1);
	R=unique(a+1,a+R+1)-a-1;
	ans=k*(k+1)/2ll;
	for(int i=1;i<=R;i++) ans-=a[i];
	cout<<ans;
}
/*

*/

D - Gomamayo Sequence

定义一下状态

f[i] [j] [k]

k表示到第几位了

i如果是0表示前面没有两两相同的，反之是1就是有

j表示最后一位结尾是什么

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int w[N];
ll f[2][2][N];
int n;
string a;
void solve()
{
cin>>n>>a;
a=" "+a;
rep(i,1,n)
{
cin>>w[i];
}
memset(f,0x3f3f3f,sizeof(f));
if(a[1]=='0')
{
f[0][1][1]=w[i];
f[0][0][1]=0;
}
else
{
f[0][0][1]=w[i];
f[0][1][1]=0;
}
rep(i,2,n)
{
if(a[i]=='0')
{
f[1][0][i]=min(f[1][1][i-1],f[0][0][i-1]);
f[0][0][i]=f[0][1][i-1];
f[0][1][i]=f[0][0][i-1]+w[i];
f[1][1][i]=w[i]+min(f[1][0][i-1],f[0][1][i-1]);
}
else
{
f[1][1][i]=min(f[1][0][i-1],f[0][1][i-1]);
f[0][1][i]=f[0][0][i-1];
f[0][0][i]=w[i]+f[0][1][i-1];
f[1][0][i]=w[i]+min(f[1][1][i-1],f[0][0][i-1]);
}
}
cout<<min(f[1][0][n],f[1][1][n]);
}
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