Codeforces Round #483 (Div. 2)

标签： codeforces

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A.Game

code

```cpp #include #include #include #include #include using namespace std; typedef long long ll; const int maxn=1000050; const int MOD=1e9+7; int a[maxn]; int main(int argc, char const *argv[]) { int n; scanf("%d", &n); for (int i = 0; i < n; ++i) { scanf("%d", a+i); } sort(a,a+n); printf("%d ", a[(n-1)/2]); return 0; } ```

## B.Minesweeper

code

```cpp #include #include #include #include #include using namespace std; const int maxn=100050; const int MOD=1e9+7; char a[150][150]; int n,m; int dx[]={0,0,1,1,1,-1,-1,-1}; int dy[]={1,-1,-1,1,0,0,1,-1}; int f(int x,int y){ int ans=0; for (int i = 0; i < 8; ++i) { int nx=x+dx[i]; int ny=y+dy[i]; if(nx>=0&&nx=0&&ny='1'&&a[i][j]<='8'&&a[i][j]-'0'!=f(i,j)) ans=false; if(a[i][j]=='.'&&f(i,j)!=0) ans=false; } } printf("%s ", ans?"Yes":"No"); return 0; } ```

## C.Finite or not?   问$frac{p}{q}$在$b$进制下能否表示为有限小数。如果可以表示为有限小数的话，我们一定可以通过把这个小数的小数点左移若干位使之成为正整数，也就是乘以若干个$b$之后变成正整数。那么我们至少需要给$p$乘以多少个$b$呢，因为$q$中质因子的最大次幂不超过$64$，所以只要$p imes b^{64}$能整除$q$即可。

code

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=300050;
typedef long long ll;
ll qmul(ll x,ll y,ll MOD){
	x%=MOD,y%=MOD;
    ll ans=(x*y-(ll)((long double)x/MOD*y+1e-3)*MOD);
    ans=(ans%MOD+MOD)%MOD;
    return ans;
}
ll Pow(ll x,ll n,ll MOD){
	ll ans=1,base=x;
	while(n){
		if(n&1) ans=qmul(ans,base,MOD);
		base=qmul(base,base,MOD);
		n>>=1;
	}
	return ans;
}
int main(int argc, char const *argv[])
{
	int q;
	scanf("%d", &q);
	while(q--){
		ll p,q,b;
		scanf("%lld%lld%lld", &p,&q,&b);
		ll a=Pow(b,64,q);
		ll ans=qmul(p,a,q);
		printf("%s
", ans==0?"Finite":"Infinite");
	}
	return 0;
}

D.XOR-pyramid

将(n)个相同元素(x)的亦或记为(x^n).
找规律可以发现$$f(a_1,a_2,a_3)=a_1 a_2^2 a_3$$

[f(a_2,a_3,a_4)=a_2 a_3^2 a_4 ]

[f(a_1,a_2,a_3,a_4)=a_1a_2^3a_3^3a4 ]

可以发现其实是一个杨辉三角,进而得出

[f(a_l,a_{l+1}......,a_{r-1},a_r)=f(a_{l+1},a_{l+2}......,a_{r-1},a_r)oplus f(a_l,a_{l+1}......,a_{r-2},a_{r-1}) ]

code

```cpp #include #include #include #include #include using namespace std; const int maxn=100050; const int MOD=1e9+7; typedef long long ll; int a[5050],n; int f[5050][5050]; int dp[5050][5050]; int main(int argc, char const *argv[]) { scanf("%d", &n); for (int i = 1; i <= n; ++i) { scanf("%d", a+i); } for (int len = 1; len <= n; ++len) { for (int i = 1; i+len-1 <= n; ++i) { int j=i+len-1; if(len==1) f[i][j]=a[i],dp[i][j]=a[i]; else f[i][j]=f[i][j-1]^f[i+1][j],dp[i][j]=max(f[i][j],max(dp[i][j-1],dp[i+1][j])); } } int q; scanf("%d", &q); while(q--){ int l,r; scanf("%d%d", &l,&r); printf("%d ", dp[l][r]); } return 0; } ```

E.Elevator

留坑