codeforces #332 div2

A.

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<map>
#include<iostream>
#include<vector>
#include<cstring>
#include<queue>
#include<string>
using namespace std;
int a,b,c;
int ans1,ans2,ans3;
int main()
{
    //freopen("input.txt","r",stdin);
    cin>>a>>b>>c;
    int ans1=2*(a+b);
    int ans2=2*(a+c);
    int ans3=2*(b+c);
    int ans4=a+b+c;
    int w=min(min(ans1,ans2),min(ans3,ans4));
    cout<<w<<endl;
}

B.

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<map>
#include<iostream>
#include<vector>
#include<cstring>
#include<queue>
#include<string>
using namespace std;
int n,m;
int ans[100005];
int w[100005];
map<int,int> q;
map<int,int>::iterator   it;
int flag=2;
int main()
{
   // freopen("input.txt","r",stdin);
    int t;
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
    {
     scanf("%d",&t);
     q.insert(pair<int,int>(t,i+1));
      ans[t]++;
    }
    for(int i=0;i<m;i++)
    {
           scanf("%d",&t);
           it=q.find(t);
           if(it!=q.end()&&flag!=1)
           {
               w[i]=it -> second;
               if(ans[it->first]>1)
                flag=0;
           }
           else
            flag=1;
    }
   if(flag==1)
    printf("Impossible
");
   else  if(flag==0)
    printf("Ambiguity
");
    else
    {
        printf("Possible
");
        for(int i=0;i<m;i++)
            printf("%d ",w[i]);
    }
}

C.

贪心：前缀最大值不大于后缀最小值

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<map>
#include<vector>
#include<iostream>
#include<cstring>
using namespace std;
int a[100005];
int ma[100005];
int mi[100005];
int main()
{
    int n,ans=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    ma[0]=-1;
    for(int i=1;i<=n;i++)
    {
        ma[i]=max(ma[i-1],a[i]);
    }
    mi[n+1]=1e9+5;
    mi[n]=a[n];
    for(int i=n-1;i>=1;i--)
    {
        mi[i]=min(mi[i+1],a[i]);
    }
    for(int i=1;i<=n;i++)
    {
        if(ma[i]<=mi[i+1])
            ans++;
    }
    printf("%d
",ans);
}

D.数学题

f(n,m)=∑k=(n−k+1)(m−k+1)

f(n,m)=(n*n/2+n/2)m+n*n+n−(n+2)(n+1)n/2+n(n+1)(2*n+1)/6

枚举n

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<map>
#include<vector>
#include<iostream>
#include<cstring>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> p;
long long n;
long long ans,x,y;
vector<p>t;

int main()
{
    scanf("%I64d",&n);
    int xx=6*pow(n,1.0/3.0);
   // printf("%d
",xx);
    for(long long i=1;i<=xx;i++)
    {
        x=(i*i+i)/2;
        y=n-i*i-i+(i+2)*(i+1)*i/2-(i+1)*i*(2*i+1)/6;
        if(y>0&&y%x==0)
        {
            ans++;
            long long dd=y/x;
            if(dd>=i){
            t.push_back(make_pair<LL,LL>(i,dd));
            t.push_back(make_pair<LL,LL>(dd,i));
            }
        }
    }
    sort(t.begin(),t.end());
    int len=unique(t.begin(),t.end())-t.begin();
    printf("%d
",len);
    for(int i=0;i<len;i++)
    {
        printf("%I64d %I64d
",t[i].first,t[i].second);
    }
}