Codeforces Round #387 (Div. 2)-优快云博客

本文解析了三道算法题目，包括寻找两个因数使乘积等于给定数值且差的绝对值最小、判断字符串中字符能否通过替换达到等数量、以及处理服务器任务分配的问题。

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题意:给出一个n 求 a*b = n 且a-b的绝对值最小

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <utility>
using namespace std;

typedef long long ll;
const int qq = 1e5 + 10;
ll num[qq];

int main(){
	ll n; cin >> n;
	int a, b;
	bool f = false;
	for(int i = 1; i <= n; ++i){
		if(n%i == 0){
			if(!f)	a = i, b = n/i, f = true;
			else{
				if(abs(i - n/i) < abs(a-b)){
					a = i, b = n/i;
				}
			}
		}
	}
	if(a >b )	swap(a, b);
	cout << a << " " << b << endl;
	return 0;
}

题意:给出一个字符串字符串, A C T G 在字符串数目中的数目要相等, ?可以变成其中任意一个, 求是否能变化后能满足条件

思路:统计一下就行

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <utility>
using namespace std;

typedef long long ll;
const int qq = 1e5 + 10;
string str;
int num[qq];

int main(){
	ll n; cin >> n;
	cin >> str;
	int maxn = 0;
	if(n%4 != 0){
		cout << "===" << endl;
		return 0;
	}
	int c = 0;
	for(int i = 0; i < (int)str.size(); ++i){
		if(str[i] == '?'){
			c++;
			continue;
		}
		num[str[i]]++;
		maxn = max(maxn, num[str[i]]);
	}
	int sum = 0;
	sum = (maxn - num['A'])+(maxn - num['C'])+(maxn - num['G'])+(maxn - num['T']);
	if(c - sum >= 0 && (c-sum)%4 == 0){
		int A = maxn - num['A'] + (c-sum)/4;
		int T = maxn - num['T'] + (c-sum)/4;
		int G = maxn - num['G'] + (c-sum)/4;
		int C = maxn - num['C'] + (c-sum)/4;
		for(int i = 0; i < (int)str.size(); ++i)
			if(str[i] == '?'){
				if(A)	str[i]='A', A--;
				else if(T)	str[i]='T', T--;
				else if(G)	str[i]='G', G--;
				else if(C)	str[i]='C', C--;
			}
		cout << str << endl;
	}else	cout << "===" << endl;
	return 0;
}

题意:有n个服务器, q个任务第i个任务在 ti 时间到来, 需要ki个服务器处理这个任务, 任务花费时间 di, 任务处理条件是当任务到来时有足够的服务器就处理, 没有就忽视这个任务

问最后每个任务的情况

思路:瞎搞搞就行了

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <utility>
using namespace std;

typedef long long ll;
const int qq = 1e6 + 4000;
struct Task{
	ll s, c, t;
	int id;
	ll sum;
}task[qq];
int vis[qq];
int to[qq];
int num[110];
vector<int> v[qq];

int cmp(Task &a, Task &b){
	return a.id < b.id;
}

int main(){
	ll n, q; cin >> n >> q;
	for(int i = 0; i < q; ++i)	cin >> task[i].s >> task[i].c >> task[i].t;
	for(int i = 0; i < q; ++i)	task[i].id = i;
	memset(vis, 0, sizeof(vis));
	memset(num, 0, sizeof(num));
	memset(to, -1, sizeof(to));
	int count = 0;
	for(int i = 0; i < q; ++i)
		to[task[i].s] = count++;
	ll p = n;
	for(int i = 1; i <= 1e6; ++i){
		if(v[i].size()!=0){
			for(int j = 0; j < (int)v[i].size(); ++j)
				num[v[i][j]] = 0;
			p += v[i].size();
		}
		if(to[i] >= 0){
			if(p >= task[to[i]].c){
				p -= task[to[i]].c;
				int res = 0;
				int k = task[to[i]].c;
				for(int j = 1; j <= n && k; ++j){
					if(!num[j]){
						num[j] = 1;
						res += j;
						v[i+task[to[i]].t].push_back(j);
						k--;
					}
				}
				task[to[i]].sum = res;
			}else	task[to[i]].sum = -1;
		}
	}
	sort(task, task+q, cmp);
	for(int i = 0; i < q; ++i)
		cout << task[i].sum << endl;
	return 0;
}