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\n");

   else  if(flag==0)

    printf("Ambiguity\n");

    else

    {

        printf("Possible\n");

        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\n",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\n",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\n",len);

    for(int i=0;i<len;i++)

    {

        printf("%I64d %I64d\n",t[i].first,t[i].second);

    }

}