Codeforces Round #387 (Div. 2)

这段时间因为工作轻松一些，所以重新开始打codeforces了。

不过成绩挺让人心寒的。

A.水题，签到

#include<stdio.h>
#include<string.h>
int ab(int x,int y)
{
    if(x-y>0) return x;
    else return y;
}
int main()
{
    int n,a,b;
    scanf("%d",&n);
    a=1;
    b=n;
    for(int i=1;i<=n/2;i++)
    {
        if(n%i==0)
        {
            int t1,t2;
            t1=i;
            t2=n/i;
            if(ab(t1,t2)<ab(a,b))
            {
                a=t1<t2?t1:t2;
                b=t1<t2?t2:t1;
            }
        }
    }
    printf("%d %d\n",a,b);
    return 0;
}

B.水题，签到

#include<stdio.h>
#include<string.h>
#define N 300
int main()
{
    int n;
    char s[N];
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%s",s);
        if(n%4!=0)
        {
            printf("===\n");
            continue;
        }
        int target=n/4;
        int t1,t2,t3,t4;
        t1=t2=t3=t4=0;
        for(int i=0;s[i]!='\0';i++)
        {
            if(s[i]=='A') t1++;
            else if(s[i]=='C') t2++;
            else if(s[i]=='G') t3++;
            else if(s[i]=='T') t4++;
        }
        t1=target-t1;
        t2=target-t2;
        t3=target-t3;
        t4=target-t4;
        if(t1<0||t2<0||t3<0||t4<0)
        {
            printf("===\n");
            continue;
        }
        int cnt1,cnt2,cnt3,cnt4;
        cnt1=cnt2=cnt3=cnt4=0;
        for(int i=0;i<n;i++)
        {
            if(s[i]=='A')
            {
                ;
            }else if(s[i]=='C')
            {
                ;
            }else if(s[i]=='G')
            {
                ;
            }else if(s[i]=='T')
            {
                ;
            }else
            {
                if(cnt1<t1)
                {
                    s[i]='A';
                    cnt1++;
                }else if(cnt2<t2)
                {
                    s[i]='C';
                    cnt2++;
                }else if(cnt3<t3)
                {
                    s[i]='G';
                    cnt3++;
                }else if(cnt4<t4)
                {
                    s[i]='T';
                    cnt4++;
                }
            }
            printf("%c",s[i]);
        }
        printf("\n");
    }
    return 0;
}

C.记录服务器使用的截止时间。时限比较宽。

#include<stdio.h>
#include<string.h>
#define N 105
int server[N];
int n,q;
int check(int t,int k,int d)
{
	int count=0;
	for(int i=1;i<=n;i++)
	{
		if(server[i]<=t)
		{
			count++;
		}	
	}	
	if(count<k)
		return -1;
	int ans=0;
	count=0;
	for(int i=1;i<=n;i++)
	{
		if(server[i]<=t)
		{
			ans+=i;
			server[i]=t+d;
			count++;
		}
		if(count==k)
		{
			return ans;
		}
	}
	return ans;
}
int main()
{
	while(scanf("%d%d",&n,&q)!=EOF)
	{
		memset(server,0,sizeof(server));
		for(int i=1;i<=q;i++)
		{
			int t,k,d;
			scanf("%d%d%d",&t,&k,&d);
			int ans=check(t,k,d);
			printf("%d\n",ans);
		}
	}
	return 0;
}

D.排序，优先填中间的坑，对两头做特殊处理。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define N 200005
int day[N],minus[N];
int cmp(const void *a,const void *b)
{
	return *(int *)a-*(int *)b;
}
int main()
{
	int n,q;
	while(scanf("%d%d",&n,&q)!=EOF)
	{
		int count=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&day[i]);
			if(day[i]<0) count++;
		}
		if(count>q)
		{
			printf("-1\n");
			continue;
		}else if(count==0)
		{
			printf("0\n");
			continue;
		}
		int left=q-count;
		int k=0;
		int current=0;
		day[n+1]=-1;
		for(int i=1;i<=n+1;i++)
		{
			if(day[i]<0)
            {
				minus[k++]=i-current-1;
                current=i;
            }
		}
		qsort(minus+1,k-2,sizeof(minus[0]),cmp);
		int ans=0;
		for(int i=1;i<k-1;i++)
		{
			if(minus[i]==0) continue;
			if(left>=minus[i])
			{
				left-=minus[i];
				continue;
			}
			ans+=2;
		}
		ans+=2;
		if(minus[k-1]<=left) ans--;
		printf("%d\n",ans);
	}
	return 0;
}

E.分割字符串，建树。dfs获取树的深度，bfs按层输出。

怎么申请内存让我搞了好久。

输出那部分写的特别丑。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<malloc.h>
#define N 1000005
char s_in[N],temp[N];
char **s;
int edge[N],next[N],head[N],vis[N],stack[N];
int cnt;
struct node
{
    int id;
    int no;
};
void addEdge(int x,int y)
{
    edge[cnt]=y;next[cnt]=head[x];head[x]=cnt++;
    return ;
}
int my_spilt()
{
    struct node q[N];
    int k=1;
    int index=0;
    int h,t;
    s=(char **)malloc(sizeof(char *)*N);
    h=t=0;
    q[t].id=0;
    q[t++].no=N;
    int flag=0;
    for(int i=0;s_in[i]!='\0';i++)
    {
        if(s_in[i]!=',')
        {
            temp[index++]=s_in[i];
        }else
        {
            temp[index]='\0';
            if(!flag)
            {
                s[k]=(char *)malloc(sizeof(char *)*(index+1));
                strcpy(s[k],temp);
                q[t].id=k;
                while(q[t-1].no==0)
                    t--;
                q[t-1].no--;
                addEdge(q[t-1].id,k++);
            }else
            {
                int no=atoi(temp);
                q[t++].no=no;
            }
            flag=(flag+1)%2;
            index=0;
        }
    }
    return 0;
}
int dfs(int father,int deep)
{
    int ans=deep-1;
    for(int i=head[father];i!=-1;i=next[i])
    {
        int son=edge[i];
        if(vis[son])
            continue;
        vis[son]=1;
        int temp=dfs(son,deep+1);
        ans=temp>ans?temp:ans;
    }
    return ans;
}
int bfs()
{
    struct node q[N];
    int h,t;
    h=t=0;
    q[t].id=0;
    q[t++].no=0;
    int count=0;
    stack[count++]=-1;
    int current=0;
    while(h!=t)
    {
        struct node cur=q[h++];
        if(cur.no==current+1)
        {
            stack[count++]=-1;
            current++;
        }
        for(int i=head[cur.id];i!=-1;i=next[i])
        {
            int son=edge[i];
            if(vis[son])
                continue;
            vis[son]=1;
            q[t].id=son;
            q[t++].no=cur.no+1;
            stack[count++]=son;
        }
    }
    return count;
}
int main()
{
    while(scanf("%s",s_in)!=EOF)
    {
        memset(head,-1,sizeof(head));
        memset(next,-1,sizeof(next));
        cnt=0;
        my_spilt();
        memset(vis,0,sizeof(vis));
        printf("%d\n",dfs(0,1));
        memset(vis,0,sizeof(vis));
        int count=bfs();
        for(int i=1;i<count;i++)
        {
            if(stack[i]==-1)
            {
                for(int j=i-1;j>=0&&stack[j]!=-1;j--)
                {
                    printf("%s ",s[stack[j]]);
                }
                printf("\n");
            }
        }
    }
    return 0;
}