[20201030P | CFSPJ1] 题目和题解

题目下载链接

T1

水题。

# include <bits/stdc++.h>
using namespace std;

const int N = 100005;

int n,k;

double gpt[N],a[N],b[N];

void FIO(void)
{
    freopen("sort.in","r",stdin);
    freopen("sort.out","w",stdout);
    return;
}

int main(void)
{
    FIO();
    scanf("%d%d",&n,&k);
    for(int i = 1; i <= n; i++)
    {
        cin >> gpt[i] >> a[i];
        b[i] = gpt[i] / a[i];
    }
    sort(b + 1,b + n + 1);
    printf("%.2lf\n",b[n - k + 1]);
    return 0;
}

T2

前缀和，单次询问\(\mathcal{O}(1)\)，预处理\(\mathcal{O}(n)\)。

# include <bits/stdc++.h>
using namespace std;

int n;

const int N = 100005;

long long a[N],q[N];

void FIO(void)
{
    freopen("sum.in","r",stdin);
    freopen("sum.out","w",stdout);
    return;
}

int main(void)
{
    FIO();
    int m;
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= n; i++)
    {
        scanf("%lld",&a[i]);
    }
    for(int i = 1; i <= n; i++) q[i] = q[i - 1] + a[i];
    int l,r;
    while(m--)
    {
        scanf("%d%d",&l,&r);
        printf("%lld\n",q[r] - q[l - 1]);
    }
    return 0;
}

T3

递推（或者叫DP），设\(f_{i,j,k}\)指走到\((i,j)\)，积分为\(k\)的线路条数。

\[f_{i,j,k} = \begin{cases} 1 & i = j = 1,k = a_{1,1}\\ f_{i - 1,j,k - a_{i,j}} + f_{i,j - 1,k - a_{i,j}} & \texttt{otherwise} \end{cases} \]

你以为是\(\mathcal{O}(n^2m^2)\)？并不是，对于\((i,j)\)，其最大\(k\)为\(10(i + j - 1)\)，易得。

# include <bits/stdc++.h>
using namespace std;
const int N = 102;
int n,m,a[N][N];
int p;
int dp[N][N][2 * N * 10];
const int mod = 1e9 + 7;

int DP(void)
{
    dp[1][1][a[1][1]] = 1;
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
        {
            if(i == 1 && j == 1) continue;
            for(int k = a[i][j]; k <= (i + j - 1) * 10; k++)
            {
                dp[i][j][k] = (dp[i - 1][j][k - a[i][j]] + dp[i][j - 1][k - a[i][j]]) % mod;
                // printf("dp[%d][%d][%d] = %d\n",i,j,k,dp[i][j][k]);
            }
        }
    }
    return dp[n][m][p] % mod;
}

void FIO(void)
{
    freopen("count.in","r",stdin);
    freopen("count.out","w",stdout);
    return;
}

int main(void)
{
    FIO();
    scanf("%d%d%d",&n,&m,&p);
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
        {
            scanf("%d",&a[i][j]);
        }
    }
    printf("%d\n",DP());
    return 0;
}

T4

吐槽题面，一开始看错一堆

注意事项：

• 最后题目中描述的“如果一头奶牛以能量\(R\)着陆在了位置\(x\)”这里的\(x\)不一定是\(x_i\)上的点，是任意一个位置。
• 干草包爆炸不连续，题目中有点含糊不清，导致我\(10\)分。

看到题很显然想到二分\(R\)。考虑对于\(R\)已知，如何去判断其是否合法。

如果我们想让一个炸弹炸掉\(x_i,x_j\)的干草包（\(x_i < x_j\))，显然放在中间（即\(\frac{x_i + x_j}{2}\)）为最优，那么易得如果\(x_j - x_i \le 2R\)，那么就能炸掉\(x_i,x_j\)和他们中间的所有干草包。

于是，题目变成了用前后差\(\le 2R\)的区间最小覆盖（？怎么感觉说起来怪怪的

# include <bits/stdc++.h>
using namespace std;
const int N = 50005;
int k;
int n;
long long x[N],c[N];

long long M = 0;

bool check(long long R)
{
    int i = 1,cnt = 0;
    while(i <= n)
    {
        int now = i + 1;
        while(now <= n && x[now] - x[i] <= 2 * R) now++;
        // if(R == 4) printf("now = %d\n",now);
        cnt++;
        if(cnt > k) return 0;
        if(now > n) break;
        i = now;
        // if(R == 4) printf("i = %d\n",i);
    }
    return cnt <= k;
}

long long qfind(void)
{
    long long l = 0,r = M;
    long long ans = 0;
    while(l <= r)
    {
        int mid = (l + r) >> 1;
        if(check(mid)) 
        {
            ans = mid;
            r = mid - 1;
        }
        else l = mid + 1;
    }
    return ans;
}

void FIO(void)
{
    freopen("angry.in","r",stdin);
    freopen("angry.out","w",stdout);
    return;
}

int main(void)
{
    FIO();
    scanf("%d%d",&n,&k);
    for(int i = 1; i <= n; i++)
    {
        scanf("%lld",&x[i]);
    }
    sort(x + 1, x + n + 1);
    M = x[n] << 1;
    printf("%lld\n",qfind());
    return 0;
}