A. Omkar and Bad Story
首先不难发现最后 (b) 数组中没有负数，因为最大的数减去一个负数后会得到一个更大的数，显然不满足题目条件。
如果 (a) 数组中有负数，则直接输出 NO，否则由于值域很小，输出 (0sim 100) 即可。
代码：
// Problem: A. Omkar and Bad Story
// Contest: Codeforces - Codeforces Round #724 (Div. 2)
// URL: https://codeforces.ml/contest/1536/problem/A
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Author: csxsl
// Created: 2021-06-27 21:31:36
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d
", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)
#define DC int T = gi <int> (); while (T--)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;
template <typename T>
inline T gi()
{
T f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return f * x;
}
const int INF = 0x3f3f3f3f, N = 303, M = N << 1;
int n, a[N];
int main()
{
//File("");
DC
{
n = gi <int> ();
for (int i = 1; i <= n; i+=1) a[i] = gi <int> ();
if (*min_element(a + 1, a + 1 + n) < 0) puts("NO");
else
{
puts("YES
101");
for (int i = 0; i <= 100; i+=1) printf("%d ", i);
puts("");
}
}
return 0;
}
B. Prinzessin der Verurteilung
一个小性质：长度为 (n) 的字符串中最多有不同的长度为 (k) 的字串 (n-k+1) 个。
那么由于 (26^2<1000<26^3)，因此直接按字典序从小到大枚举每一个长度 (le 3) 的字符串，判断它是不是这个字符串的一个子串即可。
代码：
// Problem: B. Prinzessin der Verurteilung
// Contest: Codeforces - Codeforces Round #724 (Div. 2)
// URL: https://codeforces.ml/contest/1536/problem/B
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Author: csxsl
// Created: 2021-06-27 21:40:51
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d
", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)
#define DC int T = gi <int> (); while (T--)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;
template <typename T>
inline T gi()
{
T f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return f * x;
}
const int INF = 0x3f3f3f3f, N = 100003, M = N << 1;
int n;
char s[N];
int main()
{
//File("");
DC
{
n = gi <int> (); scanf("%s", s + 1);
bool fl = false;
for (char i = 'a'; i <= 'z' && !fl; i+=1)
{
bool have = false;
for (int j = 1; j <= n; j+=1)
if (s[j] == i) have = true;
if (!have) {printf("%c
", i); fl = true; break;}
}
if (fl) continue;
for (char i = 'a'; i <= 'z' && !fl; i+=1)
for (char j = 'a'; j <= 'z' && !fl; j+=1)
{
bool have = false;
for (int k = 1; k < n; k+=1)
if (s[k] == i && s[k + 1] == j) have = true;
if (!have) {printf("%c%c
", i, j), fl = true; break;}
}
if (fl) continue;
for (char i = 'a'; i <= 'z' && !fl; i+=1)
for (char j = 'a'; j <= 'z' && !fl; j+=1)
for (char l = 'a'; l <= 'z' && !fl; l+=1)
{
bool have = false;
for (int k = 1; k < n - 1; k+=1)
if (s[k] == i && s[k + 1] == j && s[k + 2] == l) have = true;
if (!have) {printf("%c%c%c
", i, j, l), fl = true; break;}
}
}
return 0;
}
C. Diluc and Kaeya
不难发现每个子串的 (n(D):n(K)) 就等于整个字符串的 (n(D):n(K))。
设 (dp_{i,j}) 表示现在 (n(D):n(K)=i:j) 的最多划分组数，那么每加入一个字符之后这个数组的值一定是不降的，证明显然。
所以我们就可以开一个 map 记录 (dp) 数组的值，扫描字符串的时候更新并输出答案。
代码：
// Problem: C. Diluc and Kaeya
// Contest: Codeforces - Codeforces Round #724 (Div. 2)
// URL: https://codeforces.ml/contest/1536/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Author: csxsl
// Created: 2021-06-27 18:11:29
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d
", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)
#define DC int T = gi <int> (); while (T--)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;
template <typename T>
inline T gi()
{
T f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return f * x;
}
const int INF = 0x3f3f3f3f, N = 500003, M = N << 1;
int n;
char s[N];
map <PII, int> dp;
int main()
{
//File("");
DC
{
n = gi <int> ();
scanf("%s", s + 1);
dp.clear();
int sumd = 0, sumk = 0;
for (int i = 1; i <= n; i+=1)
{
if (s[i] == 'D') ++sumd;
else ++sumk;
int ld = sumd, lk = sumk;
int g = __gcd(ld, lk);
ld /= g, lk /= g;
++dp[{ld, lk}];
printf("%d ", dp[{ld, lk}]);
}
puts("");
}
return 0;
}
D. Omkar and Medians
性质：
- 每次在数列中加入两个数，排序后中位数最多移动一位；
- 如果一个长度为 (n) 的数列（(n) 为奇数）中位数为 (s)，那么至少有 (lceilfrac{n}{2} ceil) 个数 (le s)，(lceilfrac{n}{2} ceil) 个数 (ge s)。
证明可以考虑分类讨论，这里不再展开。
考虑 (b_i) 与 (b_{i-1}) 的大小关系：
- (b_i=b_{i-1})，一定可以满足条件，可以直接插入两个 (b_i)；
- (b_i<b_{i-1})，则已经有 (i-1) 个数 (<b_i)，如果之前存在一个 (b_j(j<i)) 满足 (b_i<b_j<b_{i-1})，则已经存在 (i) 个数 (<b_i)，不满足题意，直接输出
NO； - (b_i>b_{i-1})，同理找是否存在一个 (b_j(j<i)) 满足 (b_i>b_j>b_{i-1})。
整个过程可以用 set 实现，要注意细节。
// Problem: D. Omkar and Medians
// Contest: Codeforces - Codeforces Round #724 (Div. 2)
// URL: https://codeforces.ml/contest/1536/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Author: csxsl
// Created: 2021-06-27 17:54:17
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d
", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)
#define DC int T = gi <int> (); while (T--)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;
template <typename T>
inline T gi()
{
T f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return f * x;
}
const int INF = 0x3f3f3f3f, N = 200003, M = N << 1;
int n, b[N];
set <int> s;
int main()
{
//File("");
DC
{
n = gi <int> ();
for (int i = 1; i <= n; i+=1) b[i] = gi <int> ();
s.clear();
s.insert(b[1]);
bool fl = true;
for (int i = 2; i <= n; i+=1)
{
auto it1 = s.lower_bound(b[i - 1]), it2 = s.upper_bound(b[i - 1]);
bool canmove = false;
if (it1 != s.begin()) --it1, canmove = true;
if (b[i] > b[i - 1] && (it2 != s.end()) && b[i] > (*it2)) fl = false;
if (b[i] < b[i - 1] && canmove && b[i] < (*it1)) fl = false;
s.insert(b[i]);
}
puts(fl ? "YES" : "NO");
}
return 0;
}
E. Omkar and Forest
性质：如果确定每个格子填的是 (0) 还是正整数，那么存在唯一的填数方案。
证明考虑采用反证法，最后可以导出矛盾。这里不再赘述。
那么每个格子填的数要么是 (0)，要么是它与离它最近的 (0) 的曼哈顿距离。
设有 (k) 个格子为 #，那么答案为 (2^k-[k=n imes m])，因为要减去所有数都是正整数的情况。
代码：
// Problem: E. Omkar and Forest
// Contest: Codeforces - Codeforces Round #724 (Div. 2)
// URL: https://codeforces.ml/contest/1536/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Author: csxsl
// Created: 2021-06-27 18:35:57
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d
", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)
#define DC int T = gi <int> (); while (T--)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;
template <typename T>
inline T gi()
{
T f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return f * x;
}
const int INF = 0x3f3f3f3f, N = 2003, M = N << 1, mod = 1000000007;
int n, m;
char s[N][N];
int main()
{
//File("");
DC
{
n = gi <int> (), m = gi <int> ();
for (int i = 1; i <= n; i+=1) scanf("%s", s[i] + 1);
int ans = 1;
bool all = true;
for (int i = 1; i <= n; i+=1)
for (int j = 1; j <= m; j+=1)
if (s[i][j] == '0') all = false;
else ans = 1ll * ans * 2 % mod;
printf("%d
", (ans - all + mod) % mod);
}
return 0;
}
F. Omkar and Akmar
一道博弈题。
考虑最终局面，如果有偶数个格子填了数，那么后手获胜；否则先手获胜。
讨论一下可以发现最终局面只可以是填了偶数个格子。（懒得写证明了
假设有 (i) 个空格，那么有 (n-i) 个位置要填数，且 (n-i) 为偶数，只能填 BABABABA…… 或者 ABABABAB…… 的样式。
性质：相邻数之间只能有 (0) 或 (1) 个空格。证明显然。
那么最多有 (lfloorfrac{n}{2} floor) 个空格，方案数可以使用隔板法求解，为 (inom{n-i}{i})。
确定起点之后，剩下的 (n-i-1) 个格子填充顺序任意，有 ((n-i-1)!) 中方案。
起点有 (n) 中选法，可以选择 (2) 种填充样式，因此最后答案为
[2nsumlimits_{i=0}^{lfloorfrac{n}{2}
floor}[(n-i)\%2=0]inom{n-i}{i}(n-i-1)!
]
代码：
// Problem: F. Omkar and Akmar
// Contest: Codeforces - Codeforces Round #724 (Div. 2)
// URL: https://codeforces.ml/contest/1536/problem/F
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// Author: csxsl
// Created: 2021-06-27 21:18:41
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d
", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)
#define DC int T = gi <int> (); while (T--)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;
template <typename T>
inline T gi()
{
T f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return f * x;
}
const int INF = 0x3f3f3f3f, N = 1000003, M = N << 1, mod = 1000000007;
int n;
int fac[N], invfac[N];
inline int qpow(int x, int y)
{
int res = 1;
while (y)
{
if (y & 1) res = 1ll * res * x % mod;
x = 1ll * x * x % mod, y >>= 1;
}
return res;
}
inline int C(int x, int y)
{
if (x < y) return 0;
return 1ll * fac[x] * invfac[x - y] % mod * invfac[y] % mod;
}
int main()
{
//File("");
n = gi <int> ();
for (int i = (fac[0] = 1); i <= n; i+=1) fac[i] = 1ll * fac[i - 1] * i % mod;
invfac[n] = qpow(fac[n], mod - 2);
for (int i = n - 1; ~i; i-=1) invfac[i] = 1ll * invfac[i + 1] * (i + 1) % mod;
int sum = 0;
for (int i = 0; i <= n / 2; i+=1)
if ((n - i) % 2 == 0)
sum = (1ll * sum + 1ll * C(n - i, i) * fac[n - i - 1] % mod) % mod;
printf("%d
", 1ll * sum * 2 % mod * n % mod);
return 0;
}
总结
一定要善于发现题目性质，分析题目有条理，证明严谨。（当然赛时要大胆猜结论