[Codeforces Round #513 by Barcelona Bootcamp (rated, Div. 1 + Div. 2) ](A~E)

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A:

题目大意:给你一个数字串,每个数字只可以用一次,求最多可以组成多少个电话号码(可以相同),电话号码第一个数字为$8$,且长度为$11$

题解:限制为$8$的个数和总长度,直接求

卡点:

 

C++ Code:

#include <cstdio>
#include <algorithm>
#define maxn 1000
inline int min(int a, int b) {return a < b ? a : b;}
inline int max(int a, int b) {return a > b ? a : b;}
int Tim, n, _8;
char s[maxn];
int main() {
	scanf("%d", &n);
	scanf("%s", s + 1);
	for (int i = 1; i <= n; i++) if (s[i] == ‘8‘) _8++;
	printf("%d
", min(_8, n / 11));
	return 0;
}

  

 

 

B:

题目大意:给你一个数$x$,求两个数$a,b$,使得$a+b=n$,且价值和最大,$a$的价值定义为它各个数位上的数字和

题解:$a$尽可能多分$9$,$b$为$x-a$

卡点:

 

C++ Code:

#include <cstdio>
#include <algorithm>
inline int min(int a, int b) {return a < b ? a : b;}
inline int max(int a, int b) {return a > b ? a : b;}
long long n, a, b, len;
long long ans;
inline int count(long long a) {
	int ans = 0;
	while (a) {
		ans += a % 10;
		a /= 10;
	}
	return ans;
}
int main() {
	scanf("%I64d", &n);
	long long nn = n;
	while (nn) {
		len++;
		nn /= 10;
	}
	ans = (len - 1) * 9;
	for (int i = 1; i < len; i++) a = a * 10 + 9;
	ans += count(n - a);
	printf("%I64d
", ans);
	return 0;
}

  

 

 

C:

题目大意:给定$a_1,a_2,dots,a_n$和$b_1,b_2,dots,b_m$。$c_{i,j}=a_ib_j$。

给定$x$,求四个数$x_1,y_1,x_2,y_2$满足:

$sum_{i=x_1}^{x_2}sum_{j=y_1}^{y_2}c_{i,j}leqslant x$。

对于每个满足条件的四元组$(x_1,y_1,x_2,y_2)$,定义其价值为$(x_2-x_1+1) imes(y_2-y_1+1)$,求最大价值。

题解:
$$
sum_{i=x_1}^{x_2}sum_{j=y_1}^{y2}c_{i,j}=sum_{i=x_1}^{x_2}sum_{j=y_1}^{y2}a_ib_j=(sum_{i=x_1}^{x_2}a_i)(sum_{j=y_1}^{y_2}b_j)
$$

记录下每个$x_2-x_1+1$对应最小的$sum_{i=x_1}^{x_2}a_i$,每个$y_2-y_1+1$对应最小的$sum_{j=y_1}^{y_2}b_j$,枚举,找最大的即可。

卡点:

 

C++ Code:

#include <cstdio>
#include <algorithm>
#define maxn 2010
const long long inf = 0x3f3f3f3f3f3f3f3f;
inline long long min(long long a, long long b) {return a < b ? a : b;}
inline long long max(long long a, long long b) {return a > b ? a : b;}
int a[maxn], b[maxn];
long long sa[maxn], sb[maxn], ans, x;
long long A[maxn], B[maxn];
int n, m;
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) scanf("%d", a + i), sa[i] = sa[i - 1] + a[i];
	for (int i = 1; i <= m; i++) scanf("%d", b + i), sb[i] = sb[i - 1] + b[i];
	scanf("%I64d", &x);
	for (int i = 1; i <= n; i++) {
		A[i] = inf;
		for (int j = 1; j <= n - i + 1; j++) {
			A[i] = min(sa[i + j - 1] - sa[j - 1], A[i]);
		}
//		printf("A[%d] : %lld
", i, A[i]);
	}
	for (int i = 1; i <= m; i++) {
		B[i] = inf;
		for (int j = 1; j <= m - i + 1; j++) {
			B[i] = min(sb[i + j - 1] - sb[j - 1], B[i]);
		}
//		printf("A[%d] : %lld
", i, B[i]);
	}
	ans = 0;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			if (A[i] * B[j] <= x) ans = max(ans, i * j);
		}
	}
	printf("%I64d
", ans);
	return 0;
}

  

 

 

D:

题目大意:有$n$个人,坐成若干圈(每人面对圈中间)。第$i$个人左边要有$l_i$个空位,右边要有$r_i$个空位。问至少要多少个座位。

题解:考虑把$l_i$和$r_i$分别看成一个点,则原题变为一张二分图。要求找到一个匹配,每条边的价值为$max(l_x,r_y)+1$,使得价值最小。

也就是要求“浪费”最少,即按$l$和$r$分别排序,最小连最小,次小连次小……

卡点:前两次没猜中结论。。。

 

C++ Code:

#include <cstdio>
#include <algorithm>
#define maxn 100010
inline int min(int a, int b) {return a < b ? a : b;}
inline int max(int a, int b) {return a > b ? a : b;}
int n;
long long l[maxn], r[maxn], ans, tmpl, tmpr;
int rnkl[maxn], rnkr[maxn], totl, totr;
bool vl[maxn], vr[maxn];
inline bool cmpl(int a, int b) {return l[a] < l[b];}
inline bool cmpr(int a, int b) {return r[a] < r[b];}
long long sum;
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%I64d%I64d", l + i, r + i);
	std::sort(l + 1, l + n + 1);
	std::sort(r + 1, r + n + 1);
	for (int i = 1; i <= n; i++) ans += max(l[i], r[i]);
	printf("%I64d
", ans + n);
	return 0;
}

  

 

 

E:

题目大意:一棵树。可以在任意两个在原树中距离为$2$的点之间连一条边。求所有点对间最短路的和。

题解:

$$
egin{align*}
ans&=sumlimits_{i=1}^{n-1}sumlimits_{j=i+1}^nleftlceildfrac{dis_{i,j}}{2} ight ceil\
&=dfrac{sumlimits_{i=1}^{n-1}sumlimits_{j=i+1}^ndis_{i,j}+奇数的路径数目}{2}\
end{align*}\
$$
易证,求出来的一定是整数

卡点:没开$long;long$,被$FST$了

 

C++ Code:

#include <cstdio>
#include <algorithm>
#define maxn 200010
inline int min(int a, int b) {return a < b ? a : b;}
inline int max(int a, int b) {return a > b ? a : b;}
int n;
int head[maxn], cnt = 1;
struct Edge {
	int to, nxt;
} e[maxn << 1];
inline void add(int a, int b) {
	e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
//	e[++cnt] = (Edge) {a, head[b]}; head[b] = cnt;
}
long long ans, ansji;
int fa[maxn];
long long uji[maxn], uou[maxn], dji[maxn], dou[maxn], sz[maxn];
void dfs1(int u) {
	sz[u] = 1;
	dou[u] = 1;
	for (int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if (v != fa[u]) {
			fa[v] = u;
			dfs1(v);
			ans += 1ll * sz[v] * (n - sz[v]);
			sz[u] += sz[v];
			ansji += dji[v] * dji[u] + dou[u] * dou[v];
			dou[u] += dji[v];
			dji[u] += dou[v];
		}
	}
}
int main() {
	scanf("%d", &n);
	for (int i = 1, a, b; i < n; i++) {
		scanf("%d%d", &a, &b);
		add(a, b);
		add(b, a);
	}
	dfs1(1);
	printf("%I64d
", (ans + ansji >> 1ll));
	return 0;
}

  

 

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