Codeforces Round 824 (Div. 2) 解题报告

对应场次为 CF1735

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A. Working Week

题目大意

你接下来有 \(n(6 \le n \le 10^9)\) 个工作日 必须在第 \(n\) 天休假 你可以另外再选两天休假
设你工作的三段长度为 \(l_1, l_2, l_3\)
求 \(\max(\min(\left|l_2 - l_1\right|, \left|l_3 - l_2\right|,\left|l_1 - l_3\right|))\)

Solution

懒得打 \(\LaTeX\) 了

点击查看代码

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

inline int read() {
    int xr = 0, F = 1;
    char cr;
    while (cr = getchar(), cr < '0' || cr > '9') if (cr == '-') F = -1;
    while (cr >= '0' && cr <= '9')
        xr = (xr << 3) + (xr << 1) + (cr ^ 48), cr = getchar();
    return xr * F;
}

void write(ll x) {
    char ws[51];
    int wt = 0;
    if (x < 0) putchar('-'), x = -x;
    do {
        ws[++wt] = x % 10 + '0';
        x /= 10;
    } while (x);
    for (int i = wt; i; --i) putchar(ws[i]);
}

namespace steven24 {

int T, n;

void main() {
    T = read();
    while (T--) {
        n = read();
        write((n - 3) / 3 - 1), putchar('\n');
    }
}

}

int main() {
    steven24::main();
    return 0;
}

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B. Tea with Tangerines

tangerine 橘子

题目大意

给定 \(n(1 \le n \le 100)\) 个数 你可以操作一次把一个数变成两个相加等于它的数 求最少操作多少次能使得不存在 \((x, y)\) 使得 \(2x \le y\)

Solution

通过观察样例解释不难发现最小的那个数一定是不要裂开的
然后就可以把最小数的二倍作为基准去劈其它的数
发现比较均匀地劈一定是最优的 因为这样会让劈开的几段中最长的最短

点击查看代码

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

inline int read() {
	ll xr = 0, F = 1;
	char cr;
	while (cr = getchar(), cr < '0' || cr > '9') if (cr == '-') F = -1;
	while (cr >= '0' && cr <= '9')
		xr = (xr << 3) + (xr << 1) + (cr ^ 48), cr = getchar();
	return xr * F;
}

void write(ll x) {
	char ws[51];
	int wt = 0;
	if (x < 0) putchar('-'), x = -x;
	do {
		ws[++wt] = x % 10 + '0';
		x /= 10;
	} while (x);
	for (int i = wt; i; --i) putchar(ws[i]);
}

namespace steven24 {

const int N = 521;
int a[N];
int T, n;

void main() {
	T = read();
	while (T--) {
		n = read();
		for (int i = 1; i <= n; ++i) a[i] = read();
		int minn = *min_element(a + 1, a + 1 + n);
		minn <<= 1;
		--minn;
		int ans = 0;
		for (int i = 1; i <= n; ++i) {
			if (a[i] > minn) {
				ans += a[i] / minn - 1;
				if (a[i] % minn) ++ans;
			}
		}
		write(ans), putchar('\n');
	}
}

}

int main() {
	steven24::main();
	return 0;
}