学步园

A - Pangram :判断一个字符串中有木有出现过26个字母，不论大小写，有YES没有NO

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define pi acos(-1.0)
#define eps 1e-8
typedef long long ll;
const int inf = 0x3f3f3f3f;

char a[1000];
int cnt;
int vis[30];

int main()
{
    int n;
    while( ~scanf("%d", &n ))
    {
        scanf("%s", a);
        if( n < 26 )
        {
            cout << "NO" << endl;
            continue;
        }
        cnt = 0;
        int len = strlen( a );
        memset( vis, 0, sizeof( vis ));
        for( int i = 0; i < len; i++ )
        {
            if( a[i] <= 'z' && a[i] >= 'a' )
            {
                if( !vis[ a[i] - 'a' ] )
                {
                    cnt++;
                    vis[ a[i] -'a' ] = 1;
                }
            }
            else
            {
                if( !vis[ a[i] - 'A'] )
                {
                    cnt++;
                    vis[ a[i] - 'A' ] = 1;
                }
            }
        }
        if( cnt == 26 )
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}

B - Two Buttons 
n变到m，要么乘以2，要么减去1，的最小步数。简单bfs

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define pi acos(-1.0)
#define eps 1e-8
typedef long long ll;
const int inf = 0x3f3f3f3f;

int n, m;

struct nod{
    int x, s;
};
int vis[100010];

queue <nod> q;

int bfs( int st, int ed )
{
    while( !q.empty() )
        q.pop();
    memset( vis, 0, sizeof( vis ) );
    nod t;
    t.x = st, t.s = 0;
    vis[st] = 1;
    q.push( t );
    while( !q.empty() )
    {
        nod now = q.front();
        q.pop();
        if( now.x == ed )
        {
            return now.s;
        }
        nod nxt;
        nxt.s = now.s + 1;
        if( !vis[ now.x << 1] && ( now.x << 1 ) <= 10000 )//注意不能超过范围
        {
            nxt.x = now.x << 1;
            q.push( nxt );
            vis[nxt.x] = 1;
        }
        if( !vis[ now.x - 1 ] )
        {
            nxt.x = now.x - 1;
            if( nxt.x < 0 )
                continue;
            q.push( nxt );
            vis[nxt.x] = 1;
        }
    }
}

int main()
{
    while( ~scanf("%d%d", &n, &m))
    {
        if( n <= 0 )
        {
            cout << "0" << endl;
            continue;
        }
        if( m <= n )
        {
            printf("%d\n", n - m);
            continue;
        }
        int ans = bfs( n, m );
        cout << ans << endl;
    }
    return 0;
}

C - DNA Alignment
h函数是两个相同长度的字符串，在同位置相同字符的数量，p函数是循环移位后得到的总和，然后给出S串求使p函数最大的T串有几个。

首先得到T串里的字符就是S里面字符出现最多的字符，当有多个字符出现次数一样的话，那么该位置填哪个都没关系了，又因为有n位，所以最后结果就是出现的最大次数的字母的^n

#include <set>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
const int mod = 1e9 + 7;
typedef long long LL;
typedef pair <int, int> PLL;

char str[100110];
int num[100];

int main ()
{
    int n;
    while (~scanf("%d", &n))
    {
        int maxs = 0;
        scanf("%s", str);
        memset (num, 0, sizeof(num));
        for (int i = 0; i < n; ++i)
        {
            ++num[str[i]];
            maxs = max (maxs, num[str[i]]);
        }
        int cnt = 0;
        if (num['A'] == maxs)
        {
            ++cnt;
        }
        if (num['G'] == maxs)
        {
            ++cnt;
        }
        if (num['T'] == maxs)
        {
            ++cnt;
        }
        if (num['C'] == maxs)
        {
            ++cnt;
        }
        LL ans = 1;
        for (int i = 1; i <= n; ++i)
        {
            ans *= (LL)cnt;
            ans %= mod;
        }
        cout << ans << endl;
    }
    return 0;
}