VP Codeforces Round #691 (Div. 2)
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A. Red-Blue Shuffle
题意
有n张牌，有红蓝两个数字（个位），对于一个排列，红蓝分别组成一个n位数，哪个数字大则谁赢，相同的话打平。问最后红蓝两方谁赢的概率更大。
思路
每张牌的红蓝数字是绑定的，我们只需要统计每张牌红数字更大和蓝数字更大的个数，最后看哪种情况更多即可。
代码
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f3f3f3f3f
#define int long long
#define endl "\\n"
#define rep(pigstar,a,b) for (int pigstar = (a); pigstar <= (b); pigstar ++)
#define per(pigstar,a,b) for (int pigstar = (a); pigstar >= (b); pigstar --)
#define ReadInt(pigstar,a,b) for (int i = (a); i <= (b); i ++) cin >> pigstar[i];
#define OutInt(pigstar,a,b) for (int i = (a); i <= (b); i ++) cout << pigstar[i] << " \\n"[i==b]
#define all(pigstar) pigstar.begin(),pigstar.end()
#define rvs(pigstar) reverse(pigstar.begin(),pigstar.end())
#define PII pair<int, int>
#define VI vector<int>
#define VPII vector<PII>
#define pb push_back
#define SZ(pigstar) ((int)(pigstar).size())
#define MP make_pair
#define x first
#define y second
#define CN puts("NO")
#define CY puts("YES")
//----------------debug---------------//
#define debug(args...) string _s = #args; auto end_pos = remove(_s.begin(), _s.end(), ' '); _s.erase(end_pos, _s.end()); replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); cout << "\\n";
void err(istream_iterator<string> it)
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args)
cerr << (*it) << " = " << a << " ";
err(++it, args...);
//----------------debug---------------//
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6+7;
const int mod = 1e9+7;
const double eps = 1e-6;
int qmi(int a, int k, int p)int res = 1;while (k)if (k & 1) res = (ll)res * a % p;a = (ll)a * a % p;k >>= 1;return res;
int qpow(int a,int b)int res = 1;while(b)if(b&1) res *= a;b>>=1;a*=a;return res;
int mo(int x,int p)return x = ((x%p)+p)%p;
int gcd(int a,int b)return b?gcd(b,a%b):a;
int lowbit(int x) return (-x)&x;
int fac[maxn], invfac[maxn];
void fact_init()
fac[0] = 1;
for (int i = 1; i < maxn; i ++)
fac[i] = (fac[i - 1] * i) % mod;
invfac[maxn - 1] = qmi(fac[maxn - 1], mod - 2, mod);
for (int i = maxn - 1; i; i --)
invfac[i - 1] = (invfac[i] * i) % mod;
int C (int n, int m)
return ((fac[n] * invfac[m]) % mod * invfac[n - m]) % mod;
int dx[] = 0,0,1,-1, dy[] = 1,-1,0,0;
int T = 1,N,M,K,Q;
void solve()
cin >> N;
string a,b;
cin >> a >> b;
int num = 0;
rep(i,0,N-1)
if (a[i] != b[i])
if (a[i] > b[i])
num ++;
else num --;
if (!num) puts("EQUAL");
else
if (num < 0) puts("BLUE");
else puts("RED");
signed main()
cin >> T;
// fact_init();
for (int i = 1; i <= T; i ++) solve();
return (0-0); //<3
B. Move and Turn
题意
你当前在（0,0）你可以走n步，每次选择上下左右其中一个方向，但是相邻两步不能是同一条线，必须要垂直。即上一步走了左右，这一步必须走上下；上一步走了上下，这一步必须走左右。问n步后能到达的点有几种可能，
思路
因为上下和左右交替进行，则
-
n为偶数
设n = 2k
则上下走了k步，左右走了k步。纵向上能到达的点有k + 1个，同理横向能到达的点有k + 1个，最后由(k + 1) * (k + 1)个可能点
-
n位奇数
设n = 2k + 1
- 若上下走了2k步，左右走了2k + 1步。纵向上能到达的点有k + 1个，同理横向能到达的点有k + 2个，最后由(k + 1) * (k + 2)个可能点
- 若上下走了2k + 1步，左右走了2k步。纵向上能到达的点有k + 2个，同理横向能到达的点有k + 1个，最后由(k + 1) * (k + 2)个可能点
所以最后共有2 * (k + 1) * (k + 2) 个可能点
代码
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f3f3f3f3f
#define int long long
#define endl "\\n"
#define rep(pigstar,a,b) for (int pigstar = (a); pigstar <= (b); pigstar ++)
#define per(pigstar,a,b) for (int pigstar = (a); pigstar >= (b); pigstar --)
#define ReadInt(pigstar,a,b) for (int i = (a); i <= (b); i ++) cin >> pigstar[i];
#define OutInt(pigstar,a,b) for (int i = (a); i <= (b); i ++) cout << pigstar[i] << " \\n"[i==b]
#define all(pigstar) pigstar.begin(),pigstar.end()
#define rvs(pigstar) reverse(pigstar.begin(),pigstar.end())
#define PII pair<int, int>
#define VI vector<int>
#define VPII vector<PII>
#define pb push_back
#define SZ(pigstar) ((int)(pigstar).size())
#define MP make_pair
#define x first
#define y second
#define CN puts("NO")
#define CY puts("YES")
//----------------debug---------------//
#define debug(args...) string _s = #args; auto end_pos = remove(_s.begin(), _s.end(), ' '); _s.erase(end_pos, _s.end()); replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); cout << "\\n";
void err(istream_iterator<string> it)
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args)
cerr << (*it) << " = " << a << " ";
err(++it, args...);
//----------------debug---------------//
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6+7;
const int mod = 1e9+7;
const double eps = 1e-6;
int qmi(int a, int k, int p)int res = 1;while (k)if (k & 1) res = (ll)res * a % p;a = (ll)a * a % p;k >>= 1;return res;
int qpow(int a,int b)int res = 1;while(b)if(b&1) res *= a;b>>=1;a*=a;return res;
int mo(int x,int p)return x = ((x%p)+p)%p;
int gcd(int a,int b)return b?gcd(b,a%b):a;
int lowbit(int x) return (-x)&x;
int fac[maxn], invfac[maxn];
void fact_init()
fac[0] = 1;
for (int i = 1; i < maxn; i ++)
fac[i] = (fac[i - 1] * i) % mod;
invfac[maxn - 1] = qmi(fac[maxn - 1], mod - 2, mod);
for (int i = maxn - 1; i; i --)
invfac[i - 1] = (invfac[i] * i) % mod;
int C (int n, int m)
return ((fac[n] * invfac[m]) % mod * invfac[n - m]) % mod;
int dx[] = 0,0,1,-1, dy[] = 1,-1,0,0;
int T = 1,N,M,K,Q;
void solve()
cin >> N;
vector<int> a(N);
int ans = 0;
if (N & 1)
int y = N - 1;
y /= 2;
cout << (y Codeforces Round #691 (Div. 1)
Codeforces Round #691 (Div. 1)
Codeforces Round #691 (Div. 2)
Codeforces Round #691 (Div. 2)