Codeforces Round #166 (Div. 2)

A枚舉

#include <iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
using namespace std;
int judge(int x)
{
    int f[20],g=0;
    memset(f,0,sizeof(f));
    while(x)
    {
        g++;
        int k = x%10;
        x/=10;
        if(!f[k])
        f[k] = 1;
        else
        return 0;
    }
    return 1;
}
int main()
{
    int i,j,k,n;
    cin>>n;
    for(i = n+1 ; ;i++)
    {
        if(judge(i))
        {
            cout<<i<<endl;
            break;
        }
    }
    return 0;
}
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B 枚舉+二分

#include <iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
using namespace std;
#define N 100010
int pr[N],pd[N],ma[510][510],pg[N],g;
void prime()
{
    int i,j;
    pr[1] = 1;
     for(i = 2 ; i < N ; i++)
     {
         if(!pr[i])
         {
             for(j = i+i ; j < N ; j+=i)
             if(!pr[j])
             pr[j] = 1;
         }
     }
     for(i = 2; i < N ; i++)
         if(!pr[i])
         {
             pd[g++] = i;
         }
}
int judge(int x)
{
    int i,low  =0,high=g-1,mid = (low+high)/2;
    while(low<high)
    {
        if(pd[mid]>x)
        {
            high = mid-1;
        }
        else
        low = mid+1;
        mid = (low+high)/2;
    }
    for(i = mid ; ;i++)
    if(pd[i]>x)
    break;
    return pd[i]-x;
}
int main()
{
    int i,j,k,n,m;
    prime();
    cin>>n>>m;
    for(i = 1; i <= n ; i++)
        for(j = 1; j <= m ; j++)
        cin>>ma[i][j];
    int mi = 10*N;
    for(i = 1 ; i <= n ; i++)
    {
        int s=0;
        for(j = 1; j <= m ; j++)
        {
           if(pr[ma[i][j]]!=0)
            s+=judge(ma[i][j]);
        }
        if(s<mi)
        mi = s;
    }
    for(i = 1; i <= m ; i++)
    {
        int s = 0;
        for(j = 1; j <= n ; j++)
        {
            if(pr[ma[j][i]])
            s+=judge(ma[j][i]);
        }
        if(s<mi)
        mi = s;
    }
    cout<<mi<<endl;
    return 0;
}

 

 

C 集合元素最少爲3個 元素之間不能構成等差數列

#include <iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
#define N 1000010
using namespace std;
int num[N],f[N];
int main()
{
    int i,j,k,n,o;
    cin>>n>>k;
    if(n/k>=3)
    {
        int g = 1;
        num[1] = 1;
        num[2] = 1;
        num[5] = 1;
        f[5]  =1;
        for(i = 3; i <= 3*k-6 ; i++)
        {
            if(!f[i])
            {
                g++;
                num[i] = g;
                num[i+1] = g;
                num[i+5] = g;
                f[i] = 1;
                f[i+1] = 1;
                f[i+5] = 1;
            }
        }
        num[3*k] = k;
        num[3*k-2] = k;
        num[3*k-3] = k;
        for(i = 3*k+1 ; i <= n ; i++)
        num[i] = k;
        for(i = 1; i < n ; i++)
        cout<<num[i]<<" ";
        cout<<num[n]<<endl;
    }
    else
    cout<<"-1\n";
    return 0;
}

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