算法杂题（思维、模拟）

题目链接

文章目录

A - やめて嘘はあなたらしくないよ

The hh-index of an author is the largest hh where he has at least hh papers with citations not less than hh.
Bobo has published many papers.
Given a0,a1,a2,…,ana_0, a_1, a_2, \dots, a_{n} which means Bobo has published aia_i papers with citations exactly ii, find the hh-index of Bobo.
Input
The input consists of several test cases and is terminated by end-of-file.
The first line of each test case contains an integer nn.
The second line contains (n+1)(n+1) integers a0,a1,…,ana_0, a_1, \dots, a_n.
Output
For each test case, print an integer which denotes the result.
Constraint
1≤n≤2⋅1051 \leq n \leq 2 \cdot 10^5
0≤ai≤1090 \leq a_i \leq 10^9
The sum of nn does not exceed 250,000250,000.
Sample Input1
1 2
2
1 2 3
3
0 0 0 0
Sample Output1
2
0

#include<iostream>
#include<set>
#include<vector>

int a[1000005];
using namespace std;
int main() {
	int n;
	while (~scanf("%d", &n)) {
		for (int i = 0; i <= n; ++i)scanf("%d", &a[i]);
		int ans = 0;//表示论文数
		//从后往前贪心
		for (int i = n; i >= 0; --i) {
			ans += a[i];
			//如果不满足题目条件
			if (i <= ans) {
				printf("%d\n", i);
				break;
			}
		}
	}
	return 0;
}

B - 目を見てこれからのことを話そう

The hh-index of an author is the largest hh where he has at least hh papers with citations not less than hh.
Bobo has no papers and he is going to publish some subsequently.
If he works on a paper for xx hours, the paper will get (a⋅x)(a⋅x) citations, where aa is a known constant.
It’s clear that xx should be a positive integer.
There is also a trick – one can cite his own papers published earlier.
Given Bobo has nn working hours, find the maximum hh-index of him.
Input
The input consists of several test cases and is terminated by end-of-file.
Each test case contains two integers nn and aa.
Output
For each test case, print an integer which denotes the maximum hh-index.
Constraint
1≤n≤1091\leq n\leq 10^9
0≤a≤n0\leq a\leq n
The number of test cases does not exceed 10410^4.
Sample Input
3 0
3 1
1000000000 1000000000
Sample Output
1
2
1000000000
Hint
For the first sample, Bobo can work 33 papers for 11 hour each.
With the trick mentioned, he will get papers with citations 2,1,02, 1, 0. Thus, his hh-index is 11.
For the second sample, Bobo can work 22 papers for 11 and 22 hours respectively. He will get papers with citations 1+1,2+01 + 1, 2 + 0. Thus, his hh-index is 22。

最优解是每个论文都花1小时。仔细想一想就出来了。

#include<iostream>
using namespace std;
int main() {
	int n, a;
	while (~scanf("%d%d", &n, &a)) {
		printf("%d\n", (n + a) >> 1);
	}
	return 0;
}

F - my wish かなえたいのに

Bobo has nn tuples (a1,b1,c1),(a2,b2,c2)⋯ ,(an,bn,cn)(a_1,b_1,c_1),(a_2,b_2,c_2)\cdots ,(a_n,b_n,c_n).
He would like to find the lexicographically smallest permutation p1,p2,⋯ ,pnp_1,p_2,\cdots ,p_n
of 1,2,⋯ ,n1,2,\cdots ,n.
such that for i∈{2,3,…,n}i∈\{2,3,…,n\}
it holds that
api−1+bpi−1api−1+bpi−1+cpi−1≤api+bpiapi+bpi+cpi\frac{a_{p_{i-1}}+b_{p_{i-1}}}{a_{p_{i-1}}+b_{p_{i-1}}+c_{p_{i-1}}} \leq \frac{a_{p_{i}}+b_{p_{i}}}{a_{p_{i}}+b_{p_{i}}+c_{p_{i}}}
Input
The input consists of several test cases and is terminated by end-of-file. The first line of each test case contains an integer nn.
The ithi_{th} of the following nn lines contains 3 integers ai,bia_i, b_i and cic_i.
Output
For each test case, print nn integers p1,p2,⋯ ,pnp_1,p_2,\cdots ,p_n seperated by spaces. DO NOT print trailing spaces.
Constraint
1≤n≤1031\leq n\leq 10^3
1≤ai,bi,ci≤2×1091≤a_i,b_i,c_i≤2×10^9
The sum of nn does not exceed 10^4
Sample Input
2
1 1 1
1 1 2
2
1 1 2
1 1 1
3
1 3 1
2 2 1
3 1 1
Sample Output
2 1
1 2
1 2 3

除法化乘法避免精度损失，在优化一下乘法（卡long long上限）。

#include<iostream>
#include<set>
#include<vector>
#include<algorithm>
using namespace std;
struct Node {
	long long a, b, c;
	int node;
};
Node Data[1001];
int main() {
	int n;
	while (~scanf("%d", &n)) {
		for (int i = 1; i <= n; ++i) {
			scanf("%lld%lld%lld", &Data[i].a, &Data[i].b, &Data[i].c);
			Data[i].node = i;
		}
		sort(Data + 1, Data + n + 1, [](Node&Left, Node&Right) {
			long long&&left = (Left.a + Left.b)*(Right.c),
				&&right = (Left.c)*(Right.a + Right.b);
			return (left == right ? Left.node < Right.node:left < right);
			});
		for (int i = 1; i < n; ++i) {
			printf("%lld ", Data[i].node);
		}
		printf("%lld", Data[n].node);
		putchar('\n');
	}
	return 0;
}

G - すべては God knows…

Bobo has a string S=s1s2…sn consists of letter a, b and c.
He can transform the string by inserting or deleting substrings aa, bb and abab.
Formally, A=u∘w∘v (``∘’’ denotes string concatenation) can be transformed into A′=u∘v and vice versa where u, v are (possibly empty) strings and w∈{aa,bb,abab}.
Given the target string T=t1t2⋯tmT=t_1t_2\cdots t_m, determine if Bobo can transform the string S into T.
Input
The input consists of several test cases and is terminated by end-of-file.
The first line of each test case contains a string s1s2⋯sns_1s_2\cdots s_n.
The second line contains a string t1t2⋯tmt_1t_2\cdots t_m.
Output
For each test case, print Yes if Bobo can. Print No otherwise.
Constraint
1≤n,m≤1041≤n,m≤10^4
s1,s2,⋯ ,sn,t1,t2,⋯ ,tm∈{a,b,c}s_1,s_2,\cdots ,s_n,t_1,t_2,\cdots ,t_m∈\{a,b,c\}
The sum of n and m does not exceed 250,000.
Sample Input
ab
ba
ac
ca
a
ab
Sample Output
Yes
No
No
Hint
For the first sample, Bobo can transform as ab => aababb => babb => ba.

根据提示，ab可以转为ba，ba同理。且可以增加删除aa或bb，那么，在没有c的字符串中，只要统计两个字符串中a和b个数是否同奇同偶即可。而对于c，可以看作c将字符串分割成了若干子串。

#include<iostream>
#include<string>
#define scanf scanf_s
int ans1[10001], ans2[10001];
using namespace std;string a, b;

int main() {
	while (cin >> a >> b) {
		int t = 0;
		int ct = 0;
		for (int i = 0; i < a.size(); ++i) {
			if (a[i] == 'c') {
				ans1[ct++] = t;
			}
			else {
				t ^= a[i];
			}
		}
		ans1[ct++] = t;
		int s = 0;
		int st = 0;
		for (int i = 0; i<b.size(); ++i) {
			if (b[i] == 'c') {
				ans2[st++] = s;
			}
			else {
				s ^= b[i];
			}
		}
		ans2[st++] = s;
		bool flag = false;
		if (st == ct) {
			for (int i = 0; i < st; ++i) {
				if (ans1[i] != ans2[i]) {
					puts("No");
					flag = true;
					break;
				}
			}
			if (!flag) {
				puts("Yes");
			}
		}
		else {
			puts("No");
		}
	}
	return 0;
}

K - 傷跡なぞる

Given a,b,c,da,b,c,d find out the number of pairs of integers (x,y)(x,y) where a≤x≤b,c≤y≤da≤x≤b,c≤y≤d and x∗yx*y is a multiple of 2018.
Input
The input consists of several test cases and is terminated by end-of-file.
Each test case contains four integers a,b,c,da,b,c,d.
Output
For each test case, print an integer which denotes the result.
Constraint
1≤a≤b≤109,1≤c≤d≤1091≤a≤b≤10^9,1≤c≤d≤10^9
The number of tests cases does not exceed 10410^4.
Sample Input
1 2 1 2018
1 2018 1 2018
1 1000000000 1 1000000000
Sample Output
3
6051
1485883320325200

2018的公约数只有1*2018以及2*1009这两对，根据容斥定理很容易求出结果。

#include<iostream>
#include<set>
#include<vector>
using namespace std;
long long a, b, c, d;
int main() {
	while (~scanf("%lld%lld%lld%lld", &a, &b, &c, &d)) {
		long long left[4], right[4];

		auto fuck = [](long long Array[4],long long &left,long long &right)->void {
			Array[0] = right - left + 1;
			Array[1] = right / 2 - (left - 1) / 2;
			Array[2] = right / 1009 - (left - 1) / 1009;
			Array[3] = right / 2018 - (left - 1) / 2018;
			Array[2] -= Array[3];
			Array[1] -= Array[3];
			Array[0] -= Array[3] + Array[2] + Array[1];
		};

		fuck(left, a, b);
		fuck(right, c, d);

		const long long&&Ans =
			left[0] * right[3] +
			left[1] * (right[2] + right[3]) +
			left[2] * (right[1] + right[3]) +
			left[3] * (right[0] + right[1] + right[2] + right[3]);

		printf("%lld\n", Ans);
	}
	return 0;
}
posted on 2019-12-15 06:56  SCU_GoodGuy  阅读(147)  评论(0)    收藏  举报