Codeforces Round #295 (Div. 2) ABC

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");elseputs("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;
}

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