《Codeforces Round #689 (Div. 2, based on Zed Code Competition)》

A：显然a*kb*kc*k这样构造能满足

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 2e4 + 5;
const int M = 1e6 + 5;
const LL Mod = 1e9+7;
#define pi acos(-1)
#define INF 1e9
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline LL read(){
        LL x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
        return x*f;
    }
}
using namespace FASTIO;

char a[3];
int main()
{
    a[0] = 'a',a[1] = 'b',a[2] = 'c';
    int ca;ca = read();
    while(ca--)
    {
        int n,k;n = read(),k = read();
        int tot = 0;
        for(int i = 1;i <= n;++i)
        {
            printf("%c",a[tot]);
            if(i % k == 0) tot = (tot + 1) % 3;
        }
        printf("
");
    }
    //system("pause");
    return 0;
}

B：维护一行前缀和，然后枚举即可

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 2e4 + 5;
const int M = 1e6 + 5;
const LL Mod = 1e9+7;
#define pi acos(-1)
#define INF 1e9
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline LL read(){
        LL x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
        return x*f;
    }
}
using namespace FASTIO;

string s[505];
int sum[505][505];
int main()
{
    int ca;ca = read();
    while(ca--)
    {
        memset(sum,0,sizeof(sum));
        int n,m;n = read(),m = read();
        for(int i = 0;i < n;++i) cin >> s[i];
        for(int i = 0;i < n;++i)
            for(int j = 0;j < m;++j) sum[i][j] = (j == 0 ? 0 : sum[i][j - 1]) + (s[i][j] == '*');
        LL ans = 0;
        for(int i = 0;i < n;++i)
        {
            for(int j = 0;j < m;++j)
            {
                if(s[i][j] == '*')
                {
                    ans++;
                    int lex = i + 1,ley = j - 1;
                    int rix = i + 1,riy = j + 1;
                    while(lex < n && rix < n && ley >= 0 && riy < m)
                    {
                        int ma = sum[rix][riy] - (ley == 0 ? 0 : sum[lex][ley - 1]);
                        if(ma < riy - ley + 1) break;
                        lex++,rix++;
                        ley--,riy++;
                        ans++;
                    }
                }
            }
        }
        printf("%lld
",ans);
    }
    //system("pause");
    return 0;
}

C：容斥思想。

首先，显然是该位置后面都要是全排列，这个位置才有可能到递增全排列。

那么，最后一个满足a[i] != i的位置显然就是这个截断的点，我们考虑容斥所有不可能的概率。

那么可能概率即可得到。

正常来说，这里可以正向递推，每次让其余都无法成立即可。

但是这里的数据对于一个点的概率可能多次出现，且这个贡献每次都看成一个独立的点，所以如果正向推的话去递加这个概率。

就和正确的贡献推得不一样，所以要容斥才行。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 1e5 + 5;
const int M = 1e6 + 5;
const LL Mod = 1e9+7;
#define pi acos(-1)
#define INF 1e9
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline LL read(){
        LL x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
        return x*f;
    }
}
using namespace FASTIO;

int a[N];
double b[N];
bool vis[N];
int main()
{
    int ca;ca = read();
    while(ca--)
    {
        int n,m;n = read(),m = read();
        for(int i = 1;i <= n;++i) b[i] = -1,vis[i] = 0;
        for(int i = 1;i <= n;++i) a[i] = read();
        for(int i = n;i >= 1;--i)
        {
            if(i == n && a[i] == i) vis[i] = 1; 
            else if(i != n && a[i] == i && vis[i + 1]) vis[i] = 1;
        }
        double ans = 1.0;
        while(m--)
        {
            int x;double t;
            cin >> x >> t;
            if(x == n || vis[x + 1]) ans *= (1 - t); 
        }
        if(vis[1]) printf("1.00000000
");
        else
        {
            ans = 1 - ans;
            printf("%.10f
",ans);
        }
    }
  //  system("pause");
    return 0;
}
/*
10 5
1 3 2 4 5 6 7 8 9 10
10 0.000138
1 0.000268
5 0.000341
10 0.000223
9 0.000353
*/

D：感觉比较简单，稍微看了下就想到了。

显然位置最后都会被分类排好序，那么可以设想到这个不断分解的次数肯定不多，我们把所有情况的值都标记下来即可O（1）询问。

所以分治 + 二分来分解。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 1e5 + 5;
const int M = 1e6 + 5;
const LL Mod = 1e9+7;
#define pi acos(-1)
#define INF 1e9 + 5
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline LL read(){
        LL x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
        return x*f;
    }
}
using namespace FASTIO;

int a[N];
LL sum[N];
map<LL,int> mp;
int query(int le,int ri,int val)
{
    int L = le,r = ri,ans = le - 1;
    while(L <= r)
    {
        int mid = (L + r) >> 1;
        if(a[mid] <= val) L = mid + 1,ans = mid;
        else r = mid - 1;
    }
    return ans;
}
void solve(int L,int r)
{
    int mid = (a[L] + a[r]) / 2;
    int pos = query(L,r,mid);
    LL ma1 = sum[pos] - sum[L - 1];
    LL ma2 = sum[r] - sum[pos];
    mp[ma1] = 1,mp[ma2] = 1;
    if(a[L] == a[r]) return ;
    solve(L,pos);
    solve(pos + 1,r);
}
int main()
{
    int ca;ca = read();
    while(ca--)
    {
        mp.clear();
        memset(sum,0,sizeof(sum));
        int n,q;n = read(),q = read();
        for(int i = 1;i <= n;++i) a[i] = read();
        sort(a + 1,a + n + 1);
        for(int i = 1;i <= n;++i) sum[i] = sum[i - 1] + a[i];
        solve(1,n);
        mp[sum[n]] = 1;
        while(q--)
        {
            int x;x = read();
            printf("%s
",mp[x] ? "Yes" : "No");
        }
    }
   // system("pause");
    return 0;
}

E：

贪心考虑所有情况，然后注意取模来找循环区间。