Codeforces Round 886 (Div. 4)

还是div4写起来爽

Dashboard - Codeforces Round 886 (Div. 4) - Codeforces

A. To My Critics

题意:

给出三个数a,b,c问是否存在两个数相加大于10

题解:

#define _CRT_SECURE_NO_WARNINGS
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
typedef pair PII;
const LL N = 2e5 + 10, MOD = 998244353, INF = 0x3f3f3f3f;

void solve()
{
	int a, b, c;
	scanf("%d%d%d", &a, &b, &c);
	if (a + b >= 10 || a + c >= 10 || b + c >= 10)
		printf("YES\n");
	else
		printf("NO\n");
}
int main()
{
	int T = 1;
	scanf("%d", &T);
	while (T--)
	{
		solve();
	}
	return 0;
}

B. Ten Words of Wisdom

题意:

有n对ai,bi问在ai<=10的前提下最大的bi

题解:

#define _CRT_SECURE_NO_WARNINGS
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
typedef pair PII;
const LL N = 2e5 + 10, MOD = 998244353, INF = 0x3f3f3f3f;
void solve()
{
	int n, ans = 0, t = -1;
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
	{
		int a, b;
		scanf("%d%d", &a, &b);
		if (a <= 10 && b > t)
			ans = i, t = b;
	}
	printf("%d\n", ans);
}
int main()
{
	int T = 1;
	scanf("%d", &T);
	while (T--)
	{
		solve();
	}
	return 0;
}

C. Word on the Paper

题意:

看样例就能看出来...,在8*8的矩阵里有一个竖着的英文单词,输出它

题解:

写法贼多

#define _CRT_SECURE_NO_WARNINGS
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
typedef pair PII;
const LL N = 2e5 + 10, MOD = 998244353, INF = 0x3f3f3f3f;
void solve()
{
	for (int i = 1; i <= 8; ++i)
	{
		char ch[10], c = '.';
		scanf("%s", ch + 1);
		for (int j = 1; j <= 8; ++j)
		{
			if (ch[j] != '.')
				c = ch[j];
		}
		if (c != '.')printf("%c", c);
	}
	printf("\n");
}
int main()
{
	int T = 1;
	scanf("%d", &T);
	while (T--)
	{
		solve();
	}
	return 0;
}

D. Balanced Round

题意:

给出n个数ai和一个k,问删掉最少多少个数之后,将数组排序后能使得相邻元素之间差值不大于k

题解:

直接排序然后求一个相邻元素之间差值不大于k的最大区间长度d,n-d即为答案

#define _CRT_SECURE_NO_WARNINGS
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
typedef pair PII;
const LL N = 2e5 + 10, MOD = 998244353, INF = 0x3f3f3f3f;
int a[N];
void solve()
{
	int n, k, maxx = 1;
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; ++i)
		scanf("%d", &a[i]);
	sort(a + 1, a + 1 + n);
	for (int i = 2, t = 1; i <= n; ++i)
	{
		if (a[i] - a[i - 1] > k)t = 1;
		else ++t;
		maxx = max(maxx, t);
	}
	printf("%d\n", n - maxx);
}
int main()
{
	int T = 1;
	scanf("%d", &T);
	while (T--)
	{
		solve();
	}
	return 0;
}

E. Cardboard for Pictures

题意:

给出n个数si和一个c,求使得 eq?%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%5Cleft%20%28%20S_%7Bi%7D+w%20%5Cright%20%29%5E%7B2%7D%3D%3Dc 的w

题解:

二分(可能也能直接解二元一次方程?)。注意别爆long long就行

#define _CRT_SECURE_NO_WARNINGS
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
typedef pair PII;
const LL N = 2e5 + 10, MOD = 998244353, INF = 0x3f3f3f3f;
LL n, c, a[N];
bool get_sum(LL d)
{
	LL s = 0;
	for (int i = 1; i <= n; ++i)
	{
		s += (a[i] + 2 * d) * (a[i] + 2 * d);
		if (s > c)return 0;
	}
	return 1;
}
void solve()
{
	scanf("%lld%lld", &n, &c);
	for (int i = 1; i <= n; ++i)
		scanf("%lld", &a[i]);
	LL l = 0, r = 6e8;
	while (l < r)
	{
		int mid = l + r + 1 >> 1;
		if (get_sum(mid))
			l = mid;
		else
			r = mid - 1;
	}
	printf("%lld\n", l);
}
int main()
{
	int T = 1;
	scanf("%d", &T);
	while (T--)
	{
		solve();
	}
	return 0;
}

F. We Were Both Children

题意:

有n只青蛙,每只青蛙初始都在0位置,每秒会往前跳ai。你可以在位置1到n设置一个陷阱,陷阱会抓住经过它的所有青蛙,求你最多能抓住多少青蛙。

题解:

在位置i放置陷阱能抓住所有跳跃距离是i的因数的青蛙,对所有跳跃距离在n以下的青蛙进行计数,枚举在1到n放置陷阱的情况,对于每种情况分解因数一下就能求能抓住的青蛙数。

#define _CRT_SECURE_NO_WARNINGS
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
typedef pair PII;
const LL N = 2e5 + 10, MOD = 998244353, INF = 0x3f3f3f3f;
int cnt[N];
void solve()
{
	int n, ans = 0;
	scanf("%d", &n);
	memset(cnt, 0, sizeof cnt);
	for (int i = 1, x; i <= n; ++i)
	{
		scanf("%d", &x);
		if (x <= n)cnt[x]++;
	}
	for (int i = 1; i <= n; ++i)
	{
		int s = 0;
		for (int j = 1; j * j <= i; ++j)
		{
			if (i % j == 0)
			{
				s += cnt[j];
				if (j * j != i)
					s += cnt[i / j];
			}
		}
		ans = max(ans, s);
	}
	printf("%d\n", ans);

}
int main()
{
	int T = 1;
	scanf("%d", &T);
	while (T--)
	{
		solve();
	}
	return 0;
}

G. The Morning Star

题意:

给出n个坐标xi,yi。有一个指南针它会指向磁石,并且它只能指向正东、正南、正西、正北以及正东南、正西南、正东北、正西北,如果它的指向不是以上八个方向,它就会坏掉。问最多有多少种方案:将磁石与指南针放到两个不同的坐标使得指南针能正常工作。

题解:

正常工作的情况:两个坐标x值相等( — ),y值相等( | ),x+y相等( \ ),x-y相等( / ),开四个map分别存x,y,x+y,x-y计数即可(一种做法见代码)

#define _CRT_SECURE_NO_WARNINGS
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
typedef pair PII;
const LL N = 2e5 + 10, MOD = 998244353, INF = 0x3f3f3f3f;
mapmp[4];
void solve()
{
	for (int i = 0; i < 4; ++i)
		mp[i].clear();
	int n;
	scanf("%d", &n);
	LL ans = 0;
	for (int i = 1; i <= n; ++i)
	{
		int x, y;
		scanf("%d%d", &x, &y);
		ans += mp[0][x]++;
		ans += mp[1][y]++;
		ans += mp[2][x - y]++;
		ans += mp[3][x + y]++;
	}
	printf("%lld\n", 2 * ans);
}
int main()
{
	int T = 1;
	scanf("%d", &T);
	while (T--)
	{
		solve();
	}
	return 0;
}

H. The Third Letter

题意:

有n个人,给出m对关系ai,bi,di:ai在bi前方di距离处(di<0说明ai在bi后方)。

问能否存在一种方案能够同时满足这m对关系

题解:

写的带权并查集...感觉挺板子的。维护并查集的同时维护x距离x的根(祖先)的距离,遍历所有关系,满足则加入并查集,不满足输出NO直接return(写过食物链的话应该挺容易想到这个的)

看了眼题目的tag发现还能建图然后dfs判断...

乐,H被hack了,明天再补代码。原来是爆int被逮住了...

#define _CRT_SECURE_NO_WARNINGS
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
typedef pair PII;
const LL N = 2e5 + 10, MOD = 998244353, INF = 0x3f3f3f3f;
int a[N], b[N], d[N], p[N];
LL s[N];
int find(int x)
{
	if (p[x] != x)
	{
		int v = p[x];
		p[x] = find(p[x]);
		s[x] += s[v];
	}
	return p[x];
}
void solve()
{
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i)
		p[i] = i, s[i] = 0;
	for (int i = 1; i <= m; ++i)
		scanf("%d%d%d", &a[i], &b[i], &d[i]);
	for (int i = 1; i <= m; ++i)
	{
		int x = find(a[i]), y = find(b[i]);
		if (x == y)
		{
			if (d[i] != s[a[i]] - s[b[i]])
			{
				printf("NO\n");
				return;
			}
		}
		else
		{
			d[i] -= s[a[i]];
			s[x] = d[i];
			p[x] = b[i];
		}
	}
	printf("YES\n");

}
int main()
{
	int T = 1;
	scanf("%d", &T);
	while (T--)
	{
		solve();
	}
	return 0;
}

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