Codeforces Round #466

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A. Points on the line

题意

给定一条直线上\(n\)个点,要求去掉最少的点,使得直线上相距最远的两个点的距离\(\leq d\).

思路

枚举长度为\(d\)的区间。

Code

#include <bits/stdc++.h>
#define F(i, a, b) for (int i = (a); i < (b); ++i)
#define F2(i, a, b) for (int i = (a); i <= (b); ++i)
#define dF(i, a, b) for (int i = (a); i > (b); --i)
#define dF2(i, a, b) for (int i = (a); i >= (b); --i)
using namespace std;
typedef long long LL;
int a[110];
int main() {
    int n, d;
    scanf("%d%d", &n, &d);
    F(i, 0, n) {
        int x;
        scanf("%d", &x);
        ++a[x];
    }
    F2(i, 1, 100) a[i] += a[i-1];
    int ans = min(a[100]-a[d], a[99-d]);
    F(i, 1, 100-d) {
        int l = i, r = i+d;
        ans = min(ans, a[l-1]+a[100]-a[r]);
    }
    printf("%d\n", ans);
    return 0;
}

B. Our Tanya is Crying Out Loud

题意

初始数为\(n\),两种操作:

  1. 减一:代价为\(A\)
  2. 除以\(k\):代价为\(B\)且仅当整除时可进行此操作

求最小代价使\(n\)变为\(1\).

思路

一旦某一次除法的代价\(\gt\)减法的(等效)代价,则直接减到\(1\),否则进行除法操作。

Code

#include <bits/stdc++.h>
#define F(i, a, b) for (int i = (a); i < (b); ++i)
#define F2(i, a, b) for (int i = (a); i <= (b); ++i)
#define dF(i, a, b) for (int i = (a); i > (b); --i)
#define dF2(i, a, b) for (int i = (a); i >= (b); --i)
using namespace std;
typedef long long LL;
int main() {
    LL n,k,a,b;
    scanf("%I64d%I64d%I64d%I64d",&n,&k,&a,&b);
    LL ans = 0;
    bool flag = false;
    while (n!=1) {
        if (n%k) {
            ans += n%k*a;
            n = n/k*k;
        }
        if (n/k*(k-1)*a <= b) {
            ans += (n-1)*a;
            break;
        }
        ans += b;
        n /= k;
    }
    printf("%I64d\n", ans);
    return 0;
}

C. Phone Numbers

题意

在字符集\(\sum\)范围内,求长度为\(k\)的,比给定字符串\(s\)大的,最小的字符串。

思路

\(len=|s|\)

  1. \(len<k\),在\(s\)之后补\(k-len\)个最小的字符
  2. \(len>k\),首先将之后的多余部分截断,再将前面的部分 加一(操作与数字+1同理)。

Code

#include <bits/stdc++.h>
#define F(i, a, b) for (int i = (a); i < (b); ++i)
#define F2(i, a, b) for (int i = (a); i <= (b); ++i)
#define dF(i, a, b) for (int i = (a); i > (b); --i)
#define dF2(i, a, b) for (int i = (a); i >= (b); --i)
#define maxn 100010
using namespace std;
typedef long long LL;
char s[maxn];
int mp[256], ch[30], a[maxn];
bool vis[256];
int main() {
    int n, k, tot=0;
    scanf("%d%d", &n,&k);
    scanf("%s",s);
    int len = strlen(s);
    F(i, 0, len) {
        if (tot==26) break;
        if (!vis[s[i]]) {
            ch[tot++] = s[i];
            vis[s[i]] = true;
        }
    }
    sort(ch, ch+tot);
    F(i, 0, tot) mp[ch[i]] = i;
    if (k>len) {
        while (len<k) s[len++] = ch[0];
        s[len] = '\0';
        puts(s);
        return 0;
    }
    s[k] = '\0';
    F(i, 0, k) a[i] = mp[s[i]];
    int i = k-1;
    while (true) {
        ++a[i];
        if (a[i]<tot) break;
        a[i--] = 0;
    }
    F(i, 0, k) s[i] = ch[a[i]];
    puts(s);
    return 0;
}

D. Alena And The Heater

题意

给定数组\(a\),按一定的规则可以生成数组\(b\). 规则如下:

  • \(b_1=b_2=b_3=b_4=0.\)
  • For all \(5\leq i\leq n\):
    --- \(b_i=0\) if \(a_i,a_{i-1},a_{i-2},a_{i-3},a_{i-4}\gt r\) and \(b_{i-1},b_{i-2},b_{i-3},b_{i-4}=1\)
    --- \(b_i=1\) if \(a_i,a_{i-1},a_{i-2},a_{i-3},a_{i-4}\lt l\) and \(b_{i-1},b_{i-2},b_{i-3},b_{i-4}=0\)
    --- \(b_i=b_{i-1}\) otherwise

思路

只需确定每个转折点,对于其及其前面共\(5\)个数字,由\(1\)变为\(0\)则取最小值使\(r\)比它小,由\(0\)变为\(1\)则取最大值使\(l\)比它大。

至于中间则无需考虑,因为题目保证有解,而要不发生变化,即条件不成立,则只要\(l\)尽量小,\(r\)尽量大即可。

Code

#include <bits/stdc++.h>
#define F(i, a, b) for (int i = (a); i < (b); ++i)
#define F2(i, a, b) for (int i = (a); i <= (b); ++i)
#define dF(i, a, b) for (int i = (a); i > (b); --i)
#define dF2(i, a, b) for (int i = (a); i >= (b); --i)
#define inf 1000000000
#define maxn 100010
using namespace std;
char s[maxn];
int a[maxn];
typedef long long LL;
int minn(int p) { int ret = a[p]; F(i, p+1, p+5) ret = min(ret, a[i]); return ret; }
int maxx(int p) { int ret = a[p]; F(i, p+1, p+5) ret = max(ret, a[i]); return ret; }
int main() {
    int n;
    scanf("%d", &n);
    F(i, 0, n) scanf("%d", &a[i]);
    scanf("%s", s);
    bool flag = false;
    int r = inf, l = -inf;
    F(i, 0, n) {
        if (s[i]-'0'!=flag) {
            if (flag) r = min(r, minn(i-4)-1);
            else l = max(l, maxx(i-4)+1);
            flag = !flag;
        }
    }
    printf("%d %d\n", l, r);
    return 0;
}

E. Cashback

题意

给定一个数组,将其分割成若干段,将每一段的\(value\)求和,要求得最小值。

\(value\)定义如下:
对于长度为\(k\)的数组\(b\),其\(value\)为除了\(\lfloor\frac{k}{c}\rfloor\)小元素以外的元素的和。
例如:[3,1,6,5,2], c=2,其value为3+6+5=14.

思路

首先一个显然的事实是:假设不分段而作为一个整体能去除掉\(m\)个元素,那分段去除掉的元素个数必然\(\leq m\).

然而又有另一个显然的事实:不分段去掉\(m\)个元素之和\(sum1\),与分段去掉\(m\)个元素\(sum2\)相比,显然\(sum1\leq sum2\). 因为不分段的话去掉的是一整段的\(m\)个最小,而分段的话去掉的是\(m\)段中每段中的最小。

综合上面两个事实,应该尽量多分段,又要保证每一段分的有价值,即,最理想的情况就是每\(c\)个元素分一段。

于是\(dp\)如下:

\(dp[i] = min(dp[i-1]+a[i], dp[i-c]+sum[i-c+1..i]-min[i-c+1..i])\)

至于最小值,直接用ST表进行RMQ即可,比赛时用的是单调队列维护的...。

Code

#include <bits/stdc++.h>
#define F(i, a, b) for (int i = (a); i < (b); ++i)
#define F2(i, a, b) for (int i = (a); i <= (b); ++i)
#define dF(i, a, b) for (int i = (a); i > (b); --i)
#define dF2(i, a, b) for (int i = (a); i >= (b); --i)
#define maxn 100010
using namespace std;
typedef long long LL;
LL a[maxn], sum[maxn], dp[maxn];
int st[maxn];
int main() {
    int n, c;
    scanf("%d%d", &n, &c);
    F2(i, 1, n) {
        scanf("%I64d", &a[i]);
        sum[i] = sum[i-1] + a[i];
    }
    if (c==1) { printf("%d\n", 0); return 0; }
    dp[1] = a[1];
    int l=0, r=0; st[r++] = 1;
    F2(i, 2, n) {
        LL minn = dp[i-1]+a[i];
        while (r>l && a[st[r-1]]>=a[i]) --r;
        st[r++] = i;
        if (i-c>=0) {
            while (r>l && st[l]<=i-c) ++l;
            minn = min(minn, dp[i-c]+sum[i]-sum[i-c]-a[st[l]]);
        }
        dp[i] = minn;
    }
    printf("%I64d\n", dp[n]);
    return 0;
}

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