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Codeforces Round 470 A-C 题解

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Codeforces Round 470 A-C 题解

Codeforces Round 470 A-C

CodeForces 948A

要将S与W用D隔开，空地全填充D判断S四周是否有S即可。从一开始可避免维护边界情况，注意输入的换行符

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#include <bits/stdc++.h>

using namespace std;
int mat[505][505];

int main()
{
    memset(mat,0,sizeof(mat));
    int n,m,ans=0;
    char c;
    cin>>n>>m;
    getchar();
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=m; j++)
        {
            scanf("%c",&c);
            if(c=='S')
                mat[i][j]=1;
            else if(c=='W')
                mat[i][j]=2;
        }
        getchar();
    }
    bool isOK=true;
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=m; j++)
        {
            if(mat[i][j]==1)
                if(mat[i-1][j]==2''mat[i+1][j]==2''mat[i][j+1]==2''mat[i][j-1]==2)
                    isOK=false;

        }
    }
    if(isOK==true)
    {
        cout<<"Yes"<<endl;
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=m; j++)
            {
                if(mat[i][j]==1)
                    cout<<"S";
                else if(mat[i][j]==2)
                    cout<<"W";
                else
                    cout<<"D";

            }
            cout<<endl;
        }
    }
    else
    {
        cout<<"No"<<endl;
    }
    return 0;
}

CodeForces 948B

数学题…不明觉厉

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#include <bits/stdc++.h>

using namespace std;
const int N = 1000000 + 5;
int prime[N];
void getprime()
{
    int n = 2;
    while (n < N)
    {
        int num = n * 2;
        while (num < N)
        {
            prime[num] = n;
            num += n;
        }
        num = n + 1;
        while (prime[num] != 0 && num < N)
        {
            num++;
        }
        n = num;
    }
}
bool cmp(int x,int y)
{
    return x>y;
}
int main()
{
    getprime();
    int x2,p2,lc,minn;
    cin >> x2;
    minn=x2;
    p2 = prime[x2];
    //cout << p2 << endl;
        for(int i=1;i<=p2;i++){
            if(prime[x2 - p2 + i])
            minn = min(minn, x2 - p2 + i - prime[x2 - p2 + i] + 1);
        }
        cout << minn << endl;
    return 0;
}

CodeForces 948C

用一个升序的优先队列维护Ti-1数组的前缀和以及当Vi的和，优先队列的顶端小于当前的前缀和，就意味着当前堆不能满足融化的值，就将该堆清空，队列剩下的值维护剩下可以呗全部融化的堆，每次输出ans即可。

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#include <bits/stdc++.h>
#define maxn 100000+5
using namespace std;
long long a[maxn],b[maxn],sum[maxn],ans;
priority_queue<long long, vector<long long>, greater<long long> > q;;//优先队列小数优先
int main()
{
    sum[0]=0;
    int n;
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
        scanf("%lld",&a[i]);
    for(int i=1; i<=n; i++)
    {
        scanf("%lld",&b[i]);
        sum[i]=sum[i-1]+b[i];
    }
    for(int i=1;i<=n;i++)
    {
        ans=0;
        q.push(a[i]+sum[i-1]);
        while(!q.empty()&&q.top()<=sum[i])
        {
            //cout<<ans<<endl;
            ans+=q.top()-sum[i-1];
            q.pop();
        }
        ans+=q.size()*b[i];
        cout<<ans<<" ";
    }
    cout<<endl;
    return 0;
}