pta甲级1005-1009+cf每日水题

1005：

简单模拟，数组打表

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
const int N=110;
int a[N];
string s;
int b[N];
int sum;
string ch[N]={
    {"zero"},
    {"one"},
    {"two"},
    {"three"},
    {"four"},
    {"five"},
    {"six"},
    {"seven"},
    {"eight"},
    {"nine"}
};
int cnt;
signed main()
{
    IOS;
    getline(cin,s);
    int num=s.length();
    for(int i=0;i<s.length();i++)
    {
        a[i]=(s[i]-'0');
        sum+=a[i];
    }
    if(!sum)
    {
        cout<<ch[0];
        return 0;
    }
    else
    {
        while(sum>0)
        {
            b[cnt++]=sum%10;
            sum=sum/10;
        }
        for(int i=cnt-1;i>=0;i--)
        {
            cout<<ch[b[i]];
            if(i>0)
            cout<<" ";
        }
    }
    return 0;
}

1006

模拟+自定义字符排序

#include<bits/stdc++.h>
using namespace std;
const int N=20;
struct node{
    char s1[100];
    char s2[100];
    char s3[100];
}p[N];
bool cmp1(node a,node b)
{ 
    if( strcmp(a.s2,b.s2)>0) return false;
    else return true;
}
bool cmp2(node a,node b)
{
    if( strcmp(a.s3,b.s3)>0) return false;
    else return true;
}
int main()
{
    int n;
    cin>>n;
     for(int i=0;i<n;i++)
     {
         cin>>p[i].s1>>p[i].s2>>p[i].s3;
     }
     sort(p,p+n,cmp1);
     cout<<p[0].s1<<" ";
     sort(p,p+n,cmp2);
     cout<<p[n-1].s1; 
     return 0; 
}

1007

最大子序列求和，dp做法

转移方程是dp[i]=max(dp[i-1]+a[i]，dp[i])

并利用pre记录以i结尾的起始前驱

#include<bits/stdc++.h>
using namespace std;
#define int long long 
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
const int N=1e4+10;
int dp[N];
int pre[N]={0,1};
int sum=-1;
int id=-1;
int a[N];
int n;
signed main()
{
    IOS;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    dp[1]=a[1];
    for(int i=2;i<=n;i++)
    {
        if(dp[i-1]+a[i]>a[i])
        {
            dp[i]=dp[i-1]+a[i];
            pre[i]=pre[i-1];
        }
        else
        {
            dp[i]=a[i];
            pre[i]=i;
        }
    }
/*    sum=*max_element(dp+1,dp+1+n);
    for(int i=1;i<=n;i++)
    {
        if(dp[i]==sum)
        {
            id=i;
            break;
        }
    }*/
    for(int i=1;i<=n;i++)
    {
        if(sum<dp[i])
        {
            sum=dp[i];
            id=i;
        }
    }
    if(id<0)
    cout<<0<<" "<<a[1]<<" "<<a[n];
    else
    cout<<sum<<" "<<a[pre[id]]<<" "<<a[id];
    return 0;
}

1008

模拟

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
const int N=1e2+10;
int n;
int sum;
int a[N];
signed main()
{
    IOS;
    cin>>n;
       for(int i=1;i<=n;i++)
       {
           cin>>a[i];
       }
       sum=a[1]*6+5;
       for(int i=2;i<=n;i++)
       {
           if(a[i]>a[i-1])
           {
               sum+=(a[i]-a[i-1])*6+5;
           }
           else
           {
               sum+=abs(a[i]-a[i-1])*4+5;
           }
       }
       cout<<sum;
    return 0;
}

1009：

Polynomials系列乘法，模拟

把握对细节的 处理

#include<bits/stdc++.h>
using namespace std;
const int num=1010;
struct node
{
    int zhishu;
    double xishu;
}poly[num];
double ans[2010];
int n,m;
int cnt;
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d %lf",&poly[i].zhishu,&poly[i].xishu);
    }
    scanf("%d",&m);
    for(int i=0;i<m;i++)
    {
        int tip;
        double tmp;
        scanf("%d %lf",&tip,&tmp);
        for(int j=0;j<n;j++)
        {
            ans[tip+poly[j].zhishu]+=(tmp*poly[j].xishu);
        }
    }
    for(int i=0;i<=2000;i++)
    {
        if(ans[i]!=0.0)
        cnt++;
    }
    printf("%d",cnt);
    for(int i=2000;i>=0;i--)
    {
        if(ans[i]!=0.0)
        printf(" %d %.1f",i,ans[i]);
    }
    return 0;
}

cf:https://codeforc.es/contest/1734/problem/A

排完序贪心策略是找中间值使得操作最小，所以说贪心的策略是满足求三根火柴操作的x,x+1,x+2，

也就是隔两个进行扫描

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
const int N=310;
int t;
int n;
int a[N];
int ans;
signed main()
{
    IOS;
    cin>>t;
    while(t--)
    {
        ans=INT_MAX;
        cin>>n;
        for(int i=1;i<=n;i++)
        cin>>a[i];
        sort(a+1,a+1+n);
        for(int i=3;i<=n;i++)
        {
            ans=min(ans,a[i]-a[i-2]);
        }
        cout<<ans<<endl;
    }
    return 0;
}