hdu 4151 4146 4145

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAX = 800000;
int a[800000];
bool is(int x)
{
     int i=0,a[11];
     memset(a,0,sizeof(a));
     while(x>0)
     {
         int tmp=x%10;
         a[tmp]++;
         if(a[tmp]>=2) return false;
         x/=10;
     }
     return true;
}
int tot=0;
void init()
{
    int i;
    for(i=1;i<=10000000;i++)
    {
        if(is(i))
        {
            a[++tot]=i;
        }
    }
}
int bs(int num)
{
    int l=1,r=tot;
    int mid;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(a[mid]>=num) r=mid-1;
        else
        {
            l=mid+1;
            if(a[l]>=num) return l;
        }
    }
}
int main()
{
    int n;
    int i,j,k;
    init();
    while(scanf("%d",&n)!=EOF)
    {
        int ans;
        if(n<0) ans=0;
        else 
        {
            ans=bs(n);
        //    if(is(n))
                ans--;
        }
        printf("%d\n",ans);
    }
    return 0;
}

做傻了，加了个树状数组多此一举

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAX = 800000;
int c[800000];
int a[800000];
bool is(int x)
{
     int i=0,a[11];
     memset(a,0,sizeof(a));
     while(x>0)
     {
         int tmp=x%10;
         a[tmp]++;
         if(a[tmp]>=2) return false;
         x/=10;
     }
     return true;
}
int tot=0;
void init()
{
    int i;
    for(i=1;i<=10000000;i++)
    {
        if(is(i))
        {
        //    flag[i]=true;
            a[++tot]=i;
        }
    }
}
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int d)
{
    while(x<=MAX)
    {
        c[x]+=d;
        x+=lowbit(x);
    }
}
int sum(int x)
{
    int ans=0;
    while(x>0)
    {
        ans+=c[x];
        x-=lowbit(x);
    }
    return ans;
}
int bs(int num)
{
    int l=1,r=tot;
    int mid;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(a[mid]>=num) r=mid-1;
        else
        {
            l=mid+1;
            if(a[l]>=num) return l;
        }
    }
//    return mid;
}
int main()
{
    int n;
    int i,j,k;
    init();
    memset(c,0,sizeof(c));
    for(i=1;i<=tot;i++)
    {
        update(i,1);
    }
    while(scanf("%d",&n)!=EOF)
    {
        int ans;
        if(n<0) ans=0;
        else 
        {
            int id;
        id=bs(n);
         id--;
        //  printf("id=%d\n",id);
        ans=sum(id);
        }
        printf("%d\n",ans);
    }
    return 0;
}

 

 

4145

先排序，再枚举

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct point {
    double x,y;
}p[100010],a,b;
double dis(point a,point b)
{
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
int cmp(point x,point y)
{
    return dis(x,a)<dis(y,a);
}
double max(double a,double b)
{
    return a>b?a:b;
}
double min(double a,double b)
{
    return a<b?a:b;
}
double aa[100010];
int main()
{
    int t;
    int i,j,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        double ans=1000000000;
        sort(p+1,p+n+1,cmp);
        aa[n]=dis(p[n],b);
        for(i=n-1;i>=1;i--)
        {
            aa[i]=max(dis(p[i],b),aa[i+1]);
        //    printf("i=%d aa=%.0lf\n",i,aa[i]);
        }
        double tmp=aa[1];//printf("tmp=%.0lf\n",tmp);
        for(i=1;i<n;i++)
        {
            tmp=min(tmp,aa[i+1]+dis(p[i],a));
        }
        tmp=min(dis(p[n],a),tmp);
        printf("%.0lf\n",tmp);
    }
}

4146

记录每个点改变次数的奇偶性即可，另外还可以开挂，就快个800ms吧，呵呵

#include<stdio.h>
#include<string.h>
int num1[1010],num2[1010];
char s[1010][1010];
int get_val()
{     
    int ret(0);     
    char c;     
    while((c=getchar())==' '||c=='\n'||c=='\r');     
        ret=c-'0';     
    while((c=getchar())!=' '&&c!='\n'&&c!='\r')                 
            ret=ret*10+c-'0';     
        return ret;
}
int main()
{
    int t,cases=1;
    int i,j,n,x,y,q;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
            scanf("%s",s[i]+1);
        scanf("%d",&q);
        memset(num1,0,sizeof(num1));
        memset(num2,0,sizeof(num2));
        while(q--)
        {
        x=get_val();
        y=get_val();
            num1[x]++;
            num2[y]++;
        }
        int ans=0;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                if(s[i][j]=='w'&&(num1[i]+num2[j])%2==0)
                    ans++;
                else if(s[i][j]=='b'&&(num1[i]+num2[j])%2==1)
                    ans++;
            }
        }
        printf("Case #%d: %d\n",cases++,ans);
    }
}