Codeforces Round 89 (Rated for Div. 2) A、B、C、D

A. Shovels and Swords 题目链接

题解：分类讨论，最后一个只需要数学(n+m)/3便可

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
ll t,n,m,k,sum,num;
ll a[105];
int main()
{
 cin>>t;
 while(t--){
  cin>>n>>m;
  if(n>m) swap(n,m);
  if(m==0||n==0) {cout<<0<<endl;continue;} 
  if(m>=2*n) {cout<<n<<endl;continue;} 
  else 
  cout<<(n+m)/3<<endl;
  } 
 return 0;
 } 

B. Shuffle 题目链接
题解：令 a=b=x,然后判断a,b与区间(l,r)的交互，ak=1在其交集的区间都可取1，min(b,l),max(a,r)

#include
#define ll long long
#define maxn 2010
#define frj(i,n,k) for(long long i=2;i<=n-k;i++)
#define mc(i,q,k) modC(i,q,k)
using namespace std;
ll n,k,t,m,x,i,a,b,sum,l,r;
int main()
{
     cin>>t;
     while(t--){
      cin>>n>>x>>m;
      a=b=x;
      for(i=1;i<=m;i++)
      {
       cin>>l>>r;
       if(l>b||a>r) continue;
       a=min(a,l);
       b=max(b,r);
   }
   cout<<b-a+1<<endl;
  }
 return 0;
}

C. Palindromic Paths题目链接
题解：只需将开始的点到某个点的0和1分开记录，取最小的改变值便可。

#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
#define PI 3.141592653589
using namespace std;
LL n,k,i,j,m,ii,jj;
LL a[75],b[75];
int main()
{
 int t;
 cin>>t;
    while(t--){
     cin>>n>>m;
     LL sum=0;
     memset(a,0,sizeof a);
     memset(b,0,sizeof b);
     for(i=0;i<n;i++)
      for(j=0;j<m;j++)
      {
      cin>>k;
           if(k) a[i+j]++;
           else b[i+j]++;
  }
  for(i=0,j=n+m-2;i<j;i++,j--)
  sum+=min(a[i]+a[j],b[i]+b[j]); 
   cout<<sum<<endl; 
 }
 return 0;
}

D. Two Divisors题目链接
欧拉筛

#include 
using namespace std;
typedef long long LL;
const int maxn = 1e7 + 59;
int prime[maxn];
bool vis[maxn];
int cnt;
int ans[maxn][2];
int res[maxn][2];
void oulasai(int n)  //欧拉筛
{
    for(int i=2;i<n;i++)
    {
        if(!vis[i]) {
         res[i][0] = i;
         prime[cnt++]=i;
        }
        for(int j=0;j<cnt;j++)
        {
            if(1LL * i*prime[j]>n) break;
            vis[1LL * i*prime[j]]=true;
            res[1LL * i*prime[j]][0] = prime[j];
            if(i%prime[j]==0) break;
        }
    }
}
int main()
{
 oulasai(maxn);
 int n;
 cin>>n;
 for(int i = 1; i <= n; i++){
  int x;
  cin>>x;
  int p1 = res[x][0];
  int xx = x;
  while(xx % p1 == 0){
   xx /= p1;
  }
  if(xx == 1) ans[i][0] = ans[i][1] = -1;
  else {
   ans[i][0] = p1;
   ans[i][1] = xx;
  }
 }
 for(int i = 1; i <= n; i++){
  cout<<ans[i][0]<<" ";
 }
 cout<<endl;
 for(int i = 1; i <= n; i++){
     cout<<ans[i][1]<<" ";
 }
    cout<<endl;
 return 0;
}