数论分块

long long H(int n) {
  long long res = 0;
  int l = 1, r;
  while (l <= n) {
    r = n / (n / l);
    res += 1LL * (r - l + 1) * (n / l);
    l = r + 1;
  }
  return res;
}

int pos = 0, mx = 0, ans = 0;
for(int i=1;i<=cnt;i++){
if(i < mx) p[i] = min(p[pos*2-i],mx-i);//处理
else p[i] = 1;//否则以i为对称轴的这个串就属于长串的范围外,无法直接得到p[i]值.
while(ss[i+p[i]] == ss[i-p[i]]) p[i]++;//还要再扩展一次
if(mx < i+p[i]) mx = i+p[i], pos = i;//更新作为长串的回文串
ans = max(ans,p[i]-1);
}
}

florr.io

digdig.io

yorg.io

generals.io

#include<bits/stdc++.h>
using namespace std;
string res1[10000],res2[10000];
map<string,string> key;
int res1len=0,res2len=0;
bool cmp(string x,string y){
	srand(res2[rand()%res2len+1][res2[rand()%res2len+1].length()-1]);
	return rand()%2==1?true:false;
}
int main() {
	int cnt=1;
	while(1){
		string a[1000];
		int len=0;
		while(1){
			cin>>a[++len];
			if(getchar()=='\n') break;
		}
		if(a[len]=="END") break;
		res2len++;
		for(int i=1;i<=len;i++){
			if(key[a[i]]!=""){
				res2[res2len]+=key[a[i]];
				res2[res2len]+=" ";
				continue;
			}
			string tmp="";
			for(int j=1;j<=cnt;j++) tmp+="_";
			res2[res2len]+=tmp;
			res2[res2len]+=" ";
			res1[++res1len]+="#define "+tmp+" "+a[i];
			key[a[i]]=tmp;
			cnt++;
		}
	}
	sort(res1+1,res1+1+res1len,cmp);
	for(int i=1;i<=res1len;i++) cout<<res1[i]<<endl;
	for(int i=1;i<=res2len;i++) cout<<res2[i]<<endl;
	return 0;
}

动态规划1 [https://www.bilibili.com/video/BV1BD4y127ZZ/?spm\_id\_from=333.337.search-card.all.click&vd\_source=71163c9b2e7202f48beb52a92e96adcd](https://www.bilibili.com/video/BV1BD4y127ZZ/?spm_id_from=333.337.search-card.all.click&vd_source=71163c9b2e7202f48beb52a92e96adcd)

动态规划2 [https://www.bilibili.com/video/BV1KA411n7Gy/?spm\_id\_from=333.337.search-card.all.click&vd\_source=71163c9b2e7202f48beb52a92e96adcd](https://www.bilibili.com/video/BV1KA411n7Gy/?spm_id_from=333.337.search-card.all.click&vd_source=71163c9b2e7202f48beb52a92e96adcd)

动态规划3 [https://www.bilibili.com/video/BV1E54y1e74X/?spm\_id\_from=333.337.search-card.all.click&vd\_source=71163c9b2e7202f48beb52a92e96adcd](https://www.bilibili.com/video/BV1E54y1e74X/?spm_id_from=333.337.search-card.all.click&vd_source=71163c9b2e7202f48beb52a92e96adcd)

动态规划4 [https://www.bilibili.com/video/BV1Rk4y117Tc?p=1&vd\_source=71163c9b2e7202f48beb52a92e96adcd](https://www.bilibili.com/video/BV1Rk4y117Tc?p=1&vd_source=71163c9b2e7202f48beb52a92e96adcd)

董晓算法 [https://space.bilibili.com/517494241](https://space.bilibili.com/517494241)

信息学一本通 [https://space.bilibili.com/3493086634183570](https://space.bilibili.com/3493086634183570)

SamShawCraft [https://space.bilibili.com/71488025](https://space.bilibili.com/71488025)

a兵长（浙师大集训) [https://space.bilibili.com/8198319/video?tid=36&special\_type=&pn=1&keyword=&order=pubdate](https://space.bilibili.com/8198319/video?tid=36&special_type=&pn=1&keyword=&order=pubdate)

今天你学废了么 [https://space.bilibili.com/3461575069403560/video?tid=0&special\_type=&pn=1&keyword=&order=pubdate](https://space.bilibili.com/3461575069403560/video?tid=0&special_type=&pn=1&keyword=&order=pubdate)

斜率优化

#include<bits/stdc++.h>
using namespace std;
int sqr(int x){return x*x;}
int q1(int i){return ;}
int q2(int i){return ;}
int slope(int x,int y){return (double)()}
int n,m,l,r,h[500005],hh[500005];
int main()
{
cin>>n>>m;
for(int i=1;i<=n;i++)
{
cin>>h[i];
h[i]+=h[i-1];
}
hh[++r]=0;
for(int i=1;i<=n;i++)
{
while(l<r&&slope(hh[l],hh[l+1])<=2*h[i])
++l;
dp[i]=;
while(l<r&&slope(hh[r-1],hh[r])>=slope(hh[r-1],i))
--r;
hh[++r]=i;
}
}