Educational Codeforces Round 42（4.11）

A. Equator

Polycarp has created his own training plan to prepare for the programming contests. He will train for nn days, all days are numbered from 11 to nn, beginning from the first.

On the ii-th day Polycarp will necessarily solve aiai problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems.

Determine the index of day when Polycarp will celebrate the equator.

Input
The first line contains a single integer nn (1≤n≤2000001≤n≤200000) — the number of days to prepare for the programming contests.

The second line contains a sequence a1,a2,…,ana1,a2,…,an (1≤ai≤100001≤ai≤10000), where aiai equals to the number of problems, which Polycarp will solve on the ii-th day.

Output
Print the index of the day when Polycarp will celebrate the equator.
找一半，累加求和用upper_bound，注意奇数一半和偶数一半，upper_bound要减一

#include<bits/stdc++.h>
using namespace std;
int a[200005];
#define ll long long
ll sm;
ll s[200005];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        int tmp;
        scanf("%d",&tmp);
        sm+=tmp;
        s[i]=sm;
    }
    ll hf;
    if(sm&1)hf=1+sm>>1;
    else hf=sm>>1;
    ll *p=upper_bound(s+1,s+1+n,hf);
    int ans=p-s;
    while(s[ans-1]>=hf)ans--;
    printf("%d\n",ans);
    return 0;
}

　B. Students in Railway Carriage

There are nn consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.

The university team for the Olympiad consists of aa student-programmers and bb student-athletes. Determine the largest number of students from all a+ba+b students, which you can put in the railway carriage so that:

no student-programmer is sitting next to the student-programmer;
and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.

Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).

Input
The first line contain three integers nn, aa and bb (1≤n≤2⋅1051≤n≤2⋅105, 0≤a,b≤2⋅1050≤a,b≤2⋅105, a+b>0a+b>0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.

The second line contains a string with length nn, consisting of characters “.” and “*”. The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.

Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
暴力贪心

#include<bits/stdc++.h>
using namespace std;
#define ll long long
char s[300005];
int num;
int main()
{
    int n,a,b;
    scanf("%d%d%d",&n,&a,&b);
    getchar();
    scanf("%s",&s);
    int nf=0;//1:a,2:b
    for(int i=0;i<n;i++)
    {
        if(s[i]=='*'){
        nf=0;continue;}
        if(a==0&&b==0)break;
        else
        {
            if(nf==0)
            {
                if(a<b)
                {
                    b--;
                    nf=2;
                    num++;
                    //printf("b sitted,nf=%d,pos %d\n",nf,i);
                }
                else
                {
                    a--;
                    nf=1;
                    num++;
                    //printf("a sitted,nf=%d,pos %d\n",nf,i);
                }
            }
            else if(nf==1)
            {
                if(b==0)
                {
                    nf=0;
                    continue;
                }
                else
                {
                    b--;
                    nf=2;
                    num++;
                    //printf("b sitted,nf=%d,pos %d\n",nf,i);
                }
            }
            else
            {
                if(a==0)
                {
                    nf=0;
                    continue;
                }
                else
                {
                    a--;
                    nf=1;
                    num++;
                //  printf("a sitted,nf=%d,pos %d\n",nf,i);
                }
            }
        }
    }
    printf("%d",num);
    return 0;
}

　C. Make a Square

You are given a positive integer nn, written without leading zeroes (for example, the number 04 is incorrect).

In one operation you can delete any digit of the given integer so that the result remains a positive integer without leading zeros.

Determine the minimum number of operations that you need to consistently apply to the given integer nn to make from it the square of some positive integer or report that it is impossible.

An integer xx is the square of some positive integer if and only if x=y2x=y2 for some positive integer yy.

Input
The first line contains a single integer nn (1≤n≤2⋅1091≤n≤2⋅109). The number is given without leading zeroes.

Output
If it is impossible to make the square of some positive integer from nn, print -1. In the other case, print the minimal number of operations required to do it.
深搜剪枝

#include<bits/stdc++.h>
using namespace std;
int cnt;
int a[11],u,od[11];
int ver(int x)
{
    for(int i=1;i*i<=x;i++)
    {
        if (i*i==x)return 1;
    }
    return 0;
}
int ans;
void dfs(int num,int last,int ord)
{
    if(ord<=ans)
    {
        for(int i=last+1;i<=u;i++)
        {
        if(ord==0&&a[i]==0)continue;
        dfs(10*num+a[i],i,ord+1);
        }
        return;
    }
    if(ver(num)){
    ans=max(ans,ord);od[ord]=1;
    }
    if(last==u)
    {
        return;
    }
    for(int i=last+1;i<=u;i++)
    {
        if(ord==0&&a[i]==0)continue;
        dfs(10*num+a[i],i,ord+1);
    }
    return;
}
int main()
{
    int x;
    scanf("%d",&x);
    stack<int> c;
    while(x>=10){
    c.push(x%10);x/=10;u++;}
    c.push(x%10);u++;
    while(!c.empty())
    {
        int tp=c.top();
        c.pop();
        a[++cnt]=tp;
    }
    dfs(0,0,0);
    printf("%d",ans==0?-1:u-ans);
    return 0;
}