#166 (Div. 2)

A.Beautiful Year

题意：找到大于y的第一个每位都不相同的年。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define clr(x) memset(x,0,sizeof(x))
int a[4];
int main()
{
    int y;
    while (scanf("%d",&y)!=EOF)
    {
        y++;
        for (;;y++)
        {
            int x = y;
            a[0] = x/1000;
            x %= 1000;
            a[1] = x/100;
            x %= 100;
            a[2] = x/10;
            a[3] = x%10;
            sort(a,a+4);
            if (a[0]!=a[1] && a[1]!=a[2] && a[2]!=a[3])
            {
                printf("%d\n",y);
                break;
            }
        }
    }
    return 0;
}

 

B.Prime Matrix

题意：在一个n*m的格子中可以给任意的格子的数+1，问至少操作多少次可以使得其中某行或者某列全为质数。

分析：打个质数表，在判断就ok。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define clr(x) memset(x,0,sizeof(x))

const int maxn = 101005;
int p[maxn];
int g[505][505];
void init()
{
    int i,j;
    clr(p);
    p[0]=p[1]=1;
    for (i=2; i<maxn; i++)
    {
        if (p[i]==0)
        {
            for (j=i*2; j<maxn; j+=i)
                p[j] = 1;
        }
    }
}
int c[maxn];
int r[maxn];
int mi[505][505];
int main()
{
   int i, j, k, x;
    int n, m, tot;
    init();
    clr(mi);
    clr(r),clr(c);
    while (scanf("%d %d",&n,&m)!=EOF)
    {
        for (i=0; i<n; i++)
            for (j=0; j<m; j++)
            {
                scanf("%d",&g[i][j]);
                x = g[i][j];
                while (p[x])
                    x++;
                mi[i][j] = x-g[i][j];

            }

                int res = 99999999;
        for (i=0; i<n; i++)
        {
            tot = 0;
            for (j=0; j<m; j++)
                tot += mi[i][j];
            r[i] = tot;
            if (r[i]<res)
                res = r[i];
        }

        for (i=0; i<m; i++)
        {
            tot = 0;
            for (j=0; j<n; j++)
                tot += mi[j][i];
            c[i] = tot;
            if (c[i]<res)
                res = c[i];
        }


        printf("%d\n",res);
    }
    return 0;
}

 

C.Secret

题意：将一个集合分成k个集合，要求每个集合的数至少有一个大于3且不成等差数列。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define clr(x) memset(x,0,sizeof(x))

int m[1000005];
int main()
{
    int n,i, j, k;
    while (scanf("%d %d",&n,&k)!=EOF)
    {
        if (k*3>n)
        {
            printf("-1\n");
            continue;
        }
        int x = n-n%k;
        int y = x-k;
        for (j=0; j<k; j++)
        {
            for (i=j+1; i<=y; i+=k)
                m[i] = j+1;
        }
        for (i=y+1; i<x; i++)
            m[i] = i-y+1;
        m[x] = 1;
        for (i=x+1; i<=n; i++)
            m[i] = i-x;
        for (i=1; i<=n; i++)
            printf("%d%c",m[i],i==n?'\n':' ');
    }
    return 0;
}

 

D.Good Substrings

题意：找到一个字符串中，包含的坏字符不超过k个的子串的个数。

分析：字典树。

#include <cstdio>
#include <cstring>
#define clr(x) memset(x,0,sizeof(x))
struct node
{
    int count;
    struct node*next[26];
}tt[2500000];
int tot;

char s[1600];
char c[27];
int main()
{
    int i, j, k, x;
    int n, m, res, sum;
    while (scanf("%s %s %d",s,c,&k)!=EOF)
    {
        res = 0;
        tot = 0;
        node *root = &tt[++tot];
        clr(tt[tot].next);
        for (i=0; s[i]; i++)
        {
            node *p = root;
            sum = 0;
            for (j=i; s[j]; j++)
            {
                if (c[s[j]-'a']=='0')
                    sum++;
                if (sum>k)
                    break;
                x = s[j]-'a';
                if (p->next[x]==NULL){
                    res++;
                    p->next[x] = &tt[++tot];
                    clr(tt[tot].next);
                }
                p = p->next[x];
            }
        }
        printf("%d\n",res);
    }
    return 0;
}

 

#include <cstdio>
#include <cstring>
int next[2500000][26];
char s[1600];
char c[27];
int main()
{
    int i, j, k;
    int res, n, tot, sum, p, x;
    while (scanf("%s %s %d",s,c,&k)!=EOF)
    {
        res = 0;
        tot = 1;
        for (i=0; s[i]; i++)
        {
            sum = 0;
            p = 1;
            for (j=i; s[j]; j++)
            {
                if (c[s[j]-'a']=='0')
                    sum++;
                if (sum>k)
                    break;
                x = s[j]-'a';
                if (next[p][x] == 0){
                    res++;
                    next[p][x] = ++tot;
                }
                p = next[p][x];
            }
        }
        printf("%d\n",res);
    }
    return 0;
}

 

#include<iostream>
#include<algorithm>
#include <cstdlib>
#include <cstdio>
using namespace std;
long long a[3000000],k,c;
char s[1520],v[30];
int main()
{
    //freopen("Data.in","r",stdin);
    while(cin>>s>>v>>k)
    {
        c = 0;
    for(int i=0;s[i];i++){
        long long kk = k;
        long long h = 0;   
        for(int j=i; s[j]&&(v[s[j]-'a']>'0'||kk--);j++)
            a[c++]=h=(h*131)^s[j];
    }
    sort(a,a+c);
    cout<<unique(a,a+c)-a<<endl;
    }
}