Daily training of training team 20180525-DIV2

Keywords: PHP

A.2295

Find how many pairs of prime numbers are equal to n.

Violence.

#include <bits/stdc++.h>
using namespace std;
int ss(int n)
{
    int i=2,a=sqrt(n);
    for(;i<=a;i++)
        if(n%i==0) return 0;
    return 1;
}
int main()
{
    int n;
    while(scanf("%d",&n),n)
    {
        int sum=0;
        for(int i=2;i<=n/2;i++)
        {
            if(ss(i)&&ss(n-i)) sum++;
        }
        printf("%d\n",sum);
    }
    return 0;
}
A.cpp

B.4607

Give whether a number can be divided by 17.

Congruence.

#include<bits/stdc++.h>
using namespace std;
int main()
{
    string s;
    while(cin>>s,s!="0")
    {
        int su=0;
        for(int i=0; s[i]; i++)su=(su*10+s[i]-'0')%17;
        cout<<(su?0:1)<<endl;
    }
}
B.cpp

C.4681

Give a 4 * 4 grid, move it once, and output the result.

Violent simulation.

#include<bits/stdc++.h>
using namespace std;
int a[5][5];
void left_swap()
{
    for(int i=0;i<4;i++)
    for(int j=1;j<4;j++)
    {
        int k=j;
        while(k>=1&&a[i][k-1]==0)
        {
            swap(a[i][k],a[i][k-1]);
            k--;
        }
    }
}
void left()
{
    for(int i=0;i<4;i++)
    for(int j=0;j<3;j++)
    if(a[i][j]==a[i][j+1])
    {
        a[i][j]+=a[i][j+1];
        a[i][j+1]=0;
        left_swap();
    }
}
void right_swap()
{
    for(int i=0;i<4;i++)
    for(int j=2;j>=0;j--)
    {
        int k=j;
        while(k<=2&&a[i][k+1]==0)
        {
            swap(a[i][k],a[i][k+1]);
            k++;
        }
    }
}
void right()
{
    for(int i=0;i<4;i++)
    for(int j=3;j>=1;j--)
    if(a[i][j]==a[i][j-1])
    {
        a[i][j]+=a[i][j-1];
        a[i][j-1]=0;
        right_swap();
    }
}
void top_swap()
{
    for(int j=0;j<4;j++)
    for(int i=1;i<4;i++)
    {
        int k=i;
        while(k>=1&&a[k-1][j]==0)
        {
            swap(a[k][j],a[k-1][j]);
            k--;
        }
    }
}
void top()
{
    for(int j=0;j<4;j++)
    for(int i=0;i<3;i++)
    if(a[i+1][j]==a[i][j])
    {
        a[i][j]+=a[i+1][j];
        a[i+1][j]=0;
        top_swap();
    }
}
void buttom_swap()
{
    for(int j=0;j<4;j++)
    for(int i=2;i>=0;i--)
    {
        int k=i;
        while(k<=2&&a[k+1][j]==0)
        {
            swap(a[k][j],a[k+1][j]);
            k++;
        }
    }
}
void buttom()
{
    for(int j=0;j<4;j++)
    for(int i=3;i>=1;i--)
    if(a[i][j]==a[i-1][j])
    {
        a[i][j]+=a[i-1][j];
        a[i-1][j]=0;
        buttom_swap();
    }
}
int main()
{
    int n;
    while(cin>>n)
    {
        memset(a,0,sizeof(a));
        for(int i=0;i<4;i++)
            for(int j=0;j<4;j++)
            cin>>a[i][j];
        if(n==0)
        {
            left_swap();
            left();
        }
        else if(n==1)
        {
            right_swap();
            right();
        }
        else if(n==2)
        {
            top_swap();
            top();
        }
        else if(n==3)
        {
            buttom_swap();
            buttom();
        }
        for(int i=0;i<4;i++)
        {
            for(int j=0;j<4;j++)
            if(j==0)cout<<a[i][j];
            else cout<<" "<<a[i][j];
            cout<<endl;
        }
    }
    return 0;
}
C.cpp

D.4682

n subordinates, each of whom needs Bi minutes to hand over tasks and Ji minutes to finish the tasks. Determine a sequence to get the task done as early as possible.

Greedy, rank Ji from big to small, and then solve.

 

#include<bits/stdc++.h>
using namespace std;
#define ll long long
struct ydw{
    ll b,j;
}qu[1005];
bool cmp(ydw x,ydw y)
{
    if(x.j!=y.j)return x.j>y.j;
    return x.b>y.b;
}
int main()
{
    ll i,n,k;
    int kk=1;
    while(scanf("%I64d",&n),n!=0)
    {
        for(i=0;i<n;i++)
        {
            cin>>qu[i].b>>qu[i].j;
        }
        sort(qu,qu+n,cmp);
        ll s=0;
        for(i=0;i<n;i++)
        {
            s+=qu[i].b;
            for(k=0;k<i;k++)
            {
                qu[k].j-=qu[i].b;
            }
        }
        ll m=0;
        for(i=0;i<n;i++)m=max(m,qu[i].j);
        printf("Case %d: ",kk++);
        cout<<s+m<<endl;
    }
    return 0;
}
D.cpp

E.3476

Give n time and ask if there is a contradiction.

Violence.

#include<stdio.h>
int main(){
    int i,j,n,h,m,s,h1,m1,s1;
    while(scanf("%d",&n)!=EOF,n){
        int f=1,a[1441]={0};
        for(i=0;i<n;i++){
            scanf("%d:%d-%d:%d",&h,&m,&h1,&m1);
            s=h*60+m;s1=h1*60+m1;
            if(f){
                for(j=s;j<s1;j++){
                    if(a[j]){
                        f=0;break;
                    }
                    a[j]=1;
                }
            }
        }
        if(f)printf("no conflict\n");
        else printf("conflict\n");
    }
}
E.cpp

F.4499

For A string of AB, select A bit in operation 1, change A to B, B to A, select A K in operation 2, and change the value of [1,k].

Thinking, looking backwards.

If s[i]='B' needs to be modified, if the modified length is > = 2, then select operation 2, otherwise select operation 1.

#include<cstdio>
int n,g,i,j;
char ch;
bool s[1000005],k;
int main()
{
    scanf("%d\n",&n);
    while(i<=n)
    {
        s[++i]=(ch=getchar())=='A'?1:0;
    }
    i--;
    while(i>=1)
    {
        if(k^s[i])i--;
        else
        {
            j=-1;
            while(!(k^s[i]))i--,j++;
            if(j)k^=1;
            g++;
        }
    }
    printf("%d",g);
}
F.cpp

G.3587

Maximum common divisor of output A, B.

Euclid algorithm.

#include<stdio.h>
__int64 gcd(__int64 a,__int64 b){
    __int64 r=a%b;
    while(r){
        a=b;b=r;r=a%b;
    }
    return b;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        __int64 a,b;
        scanf("%I64d%I64d",&a,&b);
        printf("%I64d\n",gcd(a,b));    
    }
    return 0;
}
G.cpp

H.3451

There are already some roads between villages. Your job is to build some roads to connect all villages. The length of all roads is the smallest.

Minimum spanning tree. The existing row value is 0.

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define MAXN 100+5
#define MAXM 5000+5
int Map[MAXN][MAXN];
int F[MAXN];
int n,m,fx,fy,cnt;
struct edge
{
    int u;
    int v;
    int w;
}edges[MAXM];
bool cmp(edge a,edge b)
{
    return a.w<b.w;
}
int Find(int x)
{
    return F[x]==-1?x:F[x]=Find(F[x]);
}
int Kruskal()
{
    sort(edges,edges+m,cmp);
    int ans=0;
    for(int i=0;i<m;i++)
    {
        int u,v,w;
        u=edges[i].u;
        v=edges[i].v;
        w=edges[i].w;
        fx=Find(u);
        fy=Find(v);
        if(fx!=fy)
        {
            ans+=w;
            cnt++;
            F[fx]=fy;
        }
        if(cnt==n-1)break;
    }
    if(cnt<n-1)return -1;
    else return ans;
}
int main()
{
    scanf("%d",&n);
    m=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            int w;
            scanf("%d",&w);
            if(i==j||i>j)continue;//Redundant path
            edges[m].u=i;
            edges[m].v=j;
            edges[m++].w=w;
        }
    }
    memset(F,-1,sizeof(F));
    int Q,a,b;
    scanf("%d",&Q);
    for(int i=0;i<Q;i++)
    {
        scanf("%d%d",&a,&b);
        fx=Find(a);
        fy=Find(b);
        if(fx!=fy)
        {
            F[fx]=fy;
            cnt++;
        }
    }
    int ans=Kruskal();
    printf("%d\n",ans);
    return 0;
}
H.cpp

I.3675

Give three sides and ask if they are equal.

A direct sentence is fine.

#include<bits/stdc++.h>
using namespace std;

#define ll __int64
int main()
{
    int T,ca=1;
    cin>>T;
    while(T--)
    {
        ll a[3];
        cin>>a[0]>>a[1]>>a[2];
        sort(a,a+3);
        cout<<"Case #"<<ca++<<": ";
        if(a[0]+a[1]<=a[2])cout<<"invalid!\n";
        else if(a[0]==a[1]&&a[1]==a[2])cout<<"equilateral\n";
        else if(a[0]==a[1]||a[1]==a[2])cout<<"isosceles\n";
        else cout<<"scalene\n";
    }
}
I.cpp

J.2913

[1,M] and divisible by the number of a[i].

Inclusive and exclusive: if it is an odd number, then the number is ans+=m/lcm; if it is an even number, then the number is ans-=m/lcm, where lcm is the minimum common multiple of the currently selected number.

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n,m,a[10];
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=0;i<n;i++)scanf("%d",&a[i]);
        int ans=0;
        for(int i=1;i<(1<<n);i++)
        {
            int lcm=1,cnt=0;
            for(int j=0;j<n;j++)
                if(i&(1<<j))
                {
                    lcm=a[j]/__gcd(a[j],lcm)*lcm;
                    cnt++;
                }
            if(cnt&1)ans+=m/lcm;
            else ans-=m/lcm;
        }
        printf("%d\n",ans);
    }
    return 0;
}
J.cpp

Posted by ballhogjoni on Mon, 04 Nov 2019 12:32:24 -0800

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