Codeforces Round #222 (Div. 2)

The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct.

You can assume that number a is closer to numberx than number b, if|a - x| < |b - x|.

題意：兩個人先寫一個數，然後擲篩子，誰挨著近誰贏，問贏，平個有多少種情況

注意兩個人猜的數字相等的情況

#include<iostream>
using namespace std;
int numa,numb,numdraw;
int a,b;
int cntdraw()
{
    return !((a+b)%2);
}
int cnta()
{
    return a<b?((a+b)/2-numdraw):(6-(a+b)/2);
}
int cntb()
{
    return a>b?((a+b)/2-numdraw):(6-(a+b)/2);
}
int main()
{
    cin>>a>>b;
    if(a==b)
    {
        cout<<0<<' '<<6<<' '<<0<<endl;
        return 0;
    }
    numa=numb=numdraw=0;
    numdraw=cntdraw();
    numa=cnta();
    numb=cntb();
    cout<<numa<<' '<<numdraw<<' '<<numb<<endl;
}

4
9840 9920
9860 9980
9930 10020
10040 10090

1110
1100

4
9900 9850
9940 9930
10000 10020
10060 10110

1100
1100

Consider the first sample. Each semifinal has 4 participants. The results of the first semifinal are 9840, 9860, 9930, 10040. The results of the second semifinal are 9920, 9980, 10020, 10090.

• If k = 0, the finalists are determined by the time only, so players 9840, 9860, 9920 and 9930 advance to the finals.
• If k = 1, the winners from both semifinals move to the finals (with results 9840 and 9920), and the other places are determined by the time (these places go to the sportsmen who run the distance in 9860 and 9930 milliseconds).
• If k = 2, then first and second places advance from each seminfial, these are participants with results 9840, 9860, 9920 and 9980 milliseconds.

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn=1000010;
int one[maxn],two[maxn];
void print(int k,int n)
{
    for(int i=0;i<=k;i++)
    printf("1");
    for(int i=k+1;i<n;i++)
    printf("0");
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    scanf("%d%d",&one[i],&two[i]);
    int i=0,j=0,num=0;
    while(num<n)
    {
        if(one[i]<two[j])
        i++;
        else j++;
        num++;
    }
    int k=max(i-1,n/2-1);
    print(k,n);
    puts("");
    k=max(j-1,n/2-1);
    print(k,n);
    return 0;
}

3 4 2
#..#
..#.
#...

#.X#
X.#.
#...

5 4 5
#...
#.#.
.#..
...#
.#.#

#XXX
#X#.
X#..
...#
.#.#

dfs一遍把剩餘的空地標記一下。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn=550;
int n,m,k,num;
char grid[maxn][maxn];
int s,e;
int dx[]= {1,-1,0,0};
int dy[]= {0,0,1,-1};
int cnt;
void dfs(int x,int y)
{
    if(cnt>=num-k)
        return;
    grid[x][y]='X';
    for(int i=0; i<4; i++)
    {
        int tx=x+dx[i],ty=y+dy[i];
        if(tx>=0&&tx<n&&ty>=0&&ty<m&&grid[tx][ty]=='.')
        {
            cnt++;
            dfs(tx,ty);
        }
    }
}
void print()
{
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<m; j++)
            if(grid[i][j]=='.') cout<<'X';
            else if(grid[i][j]=='X')cout<<'.';
            else cout<<grid[i][j];
        cout<<endl;
    }
}
int main()
{
    scanf("%d%d%d",&n,&m,&k);
    num=0;
    getchar();
    for(int i=0; i<n; i++)
    {
        scanf("%s",grid[i]);
        //cout<<grid[i]<<endl;
        getchar();
    }
    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++)
            if(grid[i][j]=='.')
            {
                num++,s=i,e=j;
            }
    cnt=0;
    dfs(s,e);
    print();
    return 0;
}

3 4 9
1 3 1 2
2 1 3
4 3 6

YES
2 3 2 3

3 4 10
2 3 1 2
2 1 3
4 3 6

YES
1 3 1 3

3 4 9
2 3 1 2
2 1 3
4 3 6

YES
3 3 2 3

3 4 5
1 3 1 2
2 1 3
5 3 6

NO

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstdio>
using namespace std;
const int maxn=100010;
struct node
{
    int abi,pass,id;
    friend bool operator<(const node &a,const node &b)
    {
        return a.pass>b.pass;
    }
};
node stu[maxn],vir[maxn];
int N,M,S;
bool cmp(node a,node b)
{
    return a.abi>b.abi;
}
bool can(int mid)
{
    int b=1,sum=S,i=1;
    priority_queue<node> q;
    while(b<=M)
    {
        for(;i<=N;i++)
        {
            if(stu[i].abi>=vir[b].abi)//每次找比他大的入隊
            q.push(stu[i]);
            else break;
        }
        if(q.empty()) return false;
        node tmp=q.top();
        q.pop();
        if(sum<tmp.pass) return false;
        sum-=tmp.pass;
        b+=mid;//這個大的可以連續工作mid天，消滅mid個病毒
    }
    return true;
}
void put(int ans)
{
    int i=1,b=1;
    int who[maxn];
    priority_queue<node> q;
    while(b<=M)
    {
        for(;i<=N;i++)
        {
            if(stu[i].abi>=vir[b].abi)
            q.push(stu[i]);
            else break;
        }
        node tmp=q.top();
        q.pop();
        int t=min(M,b+ans-1);
        for(int k=b;k<=t;k++)
        who[vir[k].id]=tmp.id;
        if(t>=M) break;
        b+=ans;
    }
    bool first=true;
    for(int i=1;i<=M;i++)
    {
        if(first){cout<<who[i];first=false;}
        else cout<<" "<<who[i];
    }
    cout<<endl;
}
int main()
{
    cin>>N>>M>>S;
    for(int i=1;i<=M;i++){scanf("%d",&vir[i].abi);vir[i].id=i;}
    for(int i=1;i<=N;i++){scanf("%d",&stu[i].abi);stu[i].id=i;}
    for(int i=1;i<=N;i++)scanf("%d",&stu[i].pass);
    sort(vir+1,vir+1+M,cmp);
    sort(stu+1,stu+1+N,cmp);
    if(!can(M))
    {
        cout<<"NO"<<endl;
        return 0;
    }
    int l=0,r=M,mid;
    while(l<r)
    {
        mid=(l+r)/2;
        if(can(mid)) r=mid;
        else l=mid+1;
    }
    cout<<"YES"<<endl;
    put(l);
    return 0;
}