Educational Codeforces Round 83 (Rated for Div. 2)

A - Two Regular Polygons

问是否可以在一个正多边形内部内接一个正多边形，可以内接，分配给内部的多边形的每一条边的角度一样，对应的外部多边形的边数一样

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
int Gcd(int a,int b){if (b == 0) return a; return Gcd(b , a%b);}
int Lcm(int a, int b){ return a/Gcd(a,b)*b;}
inline long long read(){
   long long f = 1, x = 0;char ch = getchar();
    while (ch > '9' || ch < '0'){if (ch == '-')f = -f;ch = getchar();}
    while (ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}
    return x * f;
}
const int maxn = 1e6 + 10;
int main(){
    int T = read();
    while(T--){
        int n = read(),m = read();
        if (n % m == 0) puts("YES");
        else puts("NO");
    }
    return 0;
}

B - Bogosort

显然，倒着放一定满足条件

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
int Gcd(int a,int b){if (b == 0) return a; return Gcd(b , a%b);}
int Lcm(int a, int b){ return a/Gcd(a,b)*b;}
inline long long read(){
   long long f = 1, x = 0;char ch = getchar();
    while (ch > '9' || ch < '0'){if (ch == '-')f = -f;ch = getchar();}
    while (ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}
    return x * f;
}
const int maxn = 1e6 + 10;
int a[maxn];
 
int main(){
    int T = read();
    while(T--){
        int n = read();
        for(int i=1; i<=n; i++){
            a[i] = read();
        }
        sort(a + 1,a + n + 1);
        for(int i=n; i>=1;i--){
            printf("%d ",a[i]);
        }
        printf("\n");
    }
    return 0;
}

C - Adding Powers

判断一下，每一个数是不是k的次幂，并且没相同次幂即可

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
int Gcd(int a,int b){if (b == 0) return a; return Gcd(b , a%b);}
int Lcm(int a, int b){ return a/Gcd(a,b)*b;}
inline long long read(){
   long long f = 1, x = 0;char ch = getchar();
    while (ch > '9' || ch < '0'){if (ch == '-')f = -f;ch = getchar();}
    while (ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}
    return x * f;
}
const int maxn = 1e6 + 10;
int cnt[maxn];
LL a[maxn];
int main(){
    int T = read();
    while(T--){
        int n = read(),k = read(),tag = 1;
        for(int i=0; i<=100; i++) cnt[i] = 0; 
        for(int i=1; i<=n; i++){
            a[i] = read();
            int pos = 0;
            LL tmp = a[i];
            while(tmp){
                if (tmp % k > 1){
                    tag = 0;
                }else{
                    if (tmp % k == 0){
                        
                    }else{
                        cnt[pos]++;
                        if (cnt[pos] > 1) tag = 0;
                    }
                }
                pos++;
                tmp /= k;
            }
        }
        if (tag) puts("YES");
        else puts("NO");
    }
    return 0;
}

D - Count the Arrays

m个里面选出n-1个，然后有n-2种情况复制一个，拿出最大的和两个复制的，剩下的n-3个数自己分配到左边还是右边，答案就是
$C(m,n-1)*(n-2)*2^{n-3}$

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
int Gcd(int a,int b){if (b == 0) return a; return Gcd(b , a%b);}
int Lcm(int a, int b){ return a/Gcd(a,b)*b;}
inline long long read(){
   long long f = 1, x = 0;char ch = getchar();
    while (ch > '9' || ch < '0'){if (ch == '-')f = -f;ch = getchar();}
    while (ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}
    return x * f;
}
const int maxn = 1e6 + 10;
const LL mod = 998244353 ;
LL C(LL n,LL m){
    static LL M = 0,inv[maxn],mul[maxn],invMul[maxn];
    while(M <= n){
        if (M){
            inv[M] = M == 1 ? 1 : (mod-mod/M)*inv[mod%M]%mod;
            mul[M] = mul[M-1]*M%mod;
            invMul[M] = invMul[M-1]*inv[M] % mod;
        }else{
            mul[M] = 1;
            invMul[M] = 1;
        }
        M++;
    }
    return mul[n]*invMul[m] % mod * invMul[n-m]%mod;
}
LL qpow(LL a,int b){
    LL ans = 1,base = a;
    while(b){
        if (b & 1) ans = ans * base % mod;
        b >>= 1;
        base = base * base % mod;
    }
    return ans % mod;
}
int main(){
    int n = read(),m = read();
    if (n == 2){
        cout << 0 << endl;
        return 0;
    }
    LL ans = C(m,n-1);
    ans = ans * (n - 2) % mod;
    ans = ans * qpow(2,n-3) % mod;
    cout << ans << endl;
    return 0;
}

E - Array Shrinking

区间dp，首先区间处理出 $[l,r]$ 区间是否可以合并成一个数，然后剩下的dp处理最短答案
$dp[l][r]=dp[l][mid]+1$
$ans[i]=min(ans[i],ans[j]+1),dp[l][r]!=0$

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
int Gcd(int a,int b){if (b == 0) return a; return Gcd(b , a%b);}
int Lcm(int a, int b){ return a/Gcd(a,b)*b;}
inline long long read(){
   long long f = 1, x = 0;char ch = getchar();
    while (ch > '9' || ch < '0'){if (ch == '-')f = -f;ch = getchar();}
    while (ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}
    return x * f;
}
const int maxn = 1e3 + 10;
int dp[maxn][maxn];
int ans[maxn];
int main(){
    int n = read();
    for(int i=1; i<=n; i++){
        dp[i][i] = read();
    }
    for(int len = 2; len <= n; len++){
        for(int l=1; l + len - 1 <= n; l++){
            int r = l + len - 1;
            for(int mid=l; mid <r; mid++){
                if (dp[l][mid] && dp[mid+1][r] && dp[l][mid] == dp[mid+1][r]){
                    dp[l][r] = dp[l][mid] + 1;
                }
            }
        }
    }
    for(int r=1; r<=n; r++){
        ans[r] = r;
        for(int l=r; l>=1; l--){
            if (dp[l][r]){
                ans[r] = min(ans[r],ans[l-1] + 1);
            }
        }
    }
    cout << ans[n] << endl;
    return 0;
}

总结

D题卡了一下，没有那么快得出思路，E题有点菜，没想出来

来源：CSDN

作者：风声sp

链接：https://blog.csdn.net/CCCCTong/article/details/104774226

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