CodeForce 148A Insomnia cure（筛法）

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我是靠谱客的博主细心香水，最近开发中收集的这篇文章主要介绍CodeForce 148A Insomnia cure（筛法），觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.

However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every k-th dragon got punched in the face with a frying pan. Every l-th dragon got his tail shut into the balcony door. Every m-th dragon got his paws trampled with sharp heels. Finally, she threatened every n-th dragon to call her mom, and he withdrew in panic.

How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of d dragons?

Input
Input data contains integer numbers k, l, m, n and d, each number in a separate line (1 ≤ k, l, m, n ≤ 10, 1 ≤ d ≤ 105).

Output
Output the number of damaged dragons.

Examples
inputCopy
1
2
3
4
12
outputCopy
12
inputCopy
2
3
4
5
24
outputCopy
17
Note
In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.

In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.

题意从第一条龙开始每k条龙（即k,2k…）会被平底锅打，每l条龙的尾巴会卡在门上，每m条龙的爪子会被踩，每n条龙会被人打电话跟他妈妈告状，问d条龙中有多少龙会要经受生理或心理上的摧残。

d数据范围到了1e5，就想着用筛法解一下

#include<iostream>
#include<sstream>
#include<string>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<unordered_set>
#include<map>
#include<unordered_map>
#include<bitset>
#include<utility>
using namespace std;
//---------------------------------------------------------
#define debug(a) cout<<#a<<" = "<<a<<endl;
#define test() cout << "============" << endl;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef vector<int> vi;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
//---------------------------------------------------------
int cnt;
const int N = 1e5 + 10;
int k, l, m, n, d;
bool st[N];
int health[N];
vi ans;
int main()
{
ios_base::sync_with_stdio(false), cin.tie(0);
cin >> k >> l >> m >> n >> d;
ans.push_back(k), ans.push_back(l), ans.push_back(m), ans.push_back(n);
sort(ans.begin(), ans.end());
if (k == 1 || l == 1 || m == 1 || n == 1)
{
cout << d << endl;
return 0;
}
for (int i = 1; i <= d; ++i)
{
for (int j = 0; j <= 3; ++j)
{
if (i == ans[j] && !st[i] )
{
st[i] = true;
for (int k = 2*i; k <= d; k += i)
{
if (!st[k])
{
st[k] = true;
}
}
}
}
}
for (int i = 1; i <= d; ++i)
{
if (st[i])
{
cnt++;
}
}
cout << cnt << endl;
return 0;
}

当然也可以有做法二简便做法：

#include<iostream>
#include<sstream>
#include<string>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<unordered_set>
#include<map>
#include<unordered_map>
#include<bitset>
#include<utility>
using namespace std;
//---------------------------------------------------------
#define debug(a) cout<<#a<<" = "<<a<<endl;
#define test() cout << "============" << endl;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef vector<int> vi;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
//---------------------------------------------------------
int main()
{
ios_base::sync_with_stdio(false), cin.tie(0);
int k,l,m,n,d;
cin>>k>>l>>m>>n>>d;
int cnt=0;
for(int i=1;i<=d;i++)
{
if(i%k==0||i%l==0||i%m==0||i%n==0) cnt++;
}
cout<<cnt<<endl;
return 0;
}