1.模拟
2.贪心
3.二分
4.数论
5.数论
6.线段树（线段树还是练少了...）
1. 蓝桥小课堂-平方和
直接模拟，注意数据范围
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl '\n'
#define int unsigned long long
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){return b > 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
void solve()
{cin >> n;cout << (n * (n + 1) * (2 * n + 1) / 6);
}
signed main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t=1;///cin>>t;while(t--){solve();}return 0;
}
2. 房顶漏水啦
贪心，分别找到最边缘的四块木板，然后考虑需要多大的正方形才能完全覆盖。
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl '\n'
#define int long long
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){return b > 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
void solve()
{cin >> n >> m;pair<int,int>ans[m];for(int i = 0 ; i < m ; i ++){cin >> ans[i].x >> ans[i].y;}int l_min = ans[0].x , l_max = ans[0].x , r_min = ans[0].y , r_max = ans[0].y;for(int i = 1 ; i < m ; i ++){l_min = min(l_min , ans[i].x);l_max = max(l_max , ans[i].x);r_min = min(r_min , ans[i].y);r_max = max(r_max , ans[i].y);}cout << max(r_max - r_min , l_max - l_min) + 1;
}
signed main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t=1;
// cin>>t;while(t--){solve();}return 0;
}
3. 质数王国
先用素数筛找素数，然后二分找出最近的素数。
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl '\n'
#define int long long
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){return b > 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
vector<LL>prime;//存储素数
bool vis[N+5];
void su()
{for(int i=2;i<=N;i++){if(!vis[i])prime.pb(i);for(int j=0;j < prime.size() && prime[j] * i <= N;j ++){vis[prime[j]*i]=1;if(i % prime[j]==0)break;}}
}
void solve()
{set<int>prim;for(auto it : prime){prim.insert(it);}cin >> n;int ans = 0;for(int i = 0 ; i < n ; i ++){int x;cin >> x;auto it = prim.lower_bound(x);int res = *it - x;if(it != prim.begin()){it--;res = min(res , x - *it);}ans += res;}cout << ans;
}
signed main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);su();int t=1;//cin>>t;while(t--){solve();}return 0;
}
4. 取余
思路：考虑遍历所有的b，看如何的去处理每一轮
显然，每一轮有的做法，就是遍历所有的a。在整个过程中会发现，
的值其实是在循环的，而每一个完整循环中
的数量是固定的，不需要逐个去计算完整循环中的
。只需要将一个完整循环中的
算出来，然后乘上循环数就行。接下来只剩下不完整的一个循环，其
的数量也是可以直接算出来的。如此便能
的去处理每一轮（注意需要特判一下S = 0 的情况，因为一轮循环中的
是
）。
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl '\n'
#define int long long
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){return b > 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
void solve()
{int a , b , s , t;cin >> a >> b >> s >> t;int ans = 0;if(s != 0){ for(int i = 1 ; i <= b ; i ++){if(i <= s){continue;}else{int lun = a / i;int res = a - lun * i;ans += lun * min(t - s + 1, i - s);if(res >= s){ans += min(t - s + 1 , res - s + 1);}}}}else{s = 1;for(int i = 1 ; i <= b ; i ++){if(i == 1){ans += a / i;}else{int lun = a / i;int res = a - lun * i;ans += lun * min(t - s + 1, i - s);ans += lun;if(res >= s){ans += min(t - s + 1 , res - s + 1);}}} }cout << ans << endl;
}
signed main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int T=1;///cin>>t;while(T--){solve();}return 0;
}
5. 数学尖子生
思路：考虑中有多少个数包含了因子
。可以想到：这些数是
的倍数。也就是说
中包含了
个包含了这些因子的数。
接下来考虑，求出来的这些数只是必要条件，要求答案必须还要找到这些数当中包含了因子的数。同样的方法，用
即可求出包含因子
的数。两数相减即答案。因为不存在数包含因子
但是不包含因子
。注意预处理一下lcm即可。（还需要注意爆longlong ， 打个表看一下）
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl '\n'
#define int long long
const LL maxn = 4e05+7;
const LL N = 1e06+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){return b > 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
int inv[N];
void solve()
{cin >> n >> m;if(n > 41){cout << 0 << endl;}elsecout << (m / inv[n - 1] - m / inv[n])<< endl;
}
signed main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t=1;inv[1] = 1;inv[0] = 1;for(int i = 2 ; i <= 41 ; i ++){int x = gcd(i , inv[i - 1]);inv[i] = lcm(i , inv[i - 1]);}cin>>t;while(t--){solve();}return 0;
}
读取配置文件全面总结)
