qining的专栏

#include#include#includeconst int maxn = 500005;typedef long long ll;int n,m;ll a[maxn],sum[maxn];struct node{ int x,y,pre,suf,dx,dy;};node tnode[maxn*2];void Build(int o,int x,int y){

本题需要对s数组做一下调整，然后求数组s中最常重复子串#include#includeconst int maxn=20005;int s[maxn],rank[maxn],height[maxn],sa[maxn],wa[maxn],wb[maxn],wv[maxn],ws[maxn],n;int mind[maxn][15],maxd[maxn][15];void adjust_

本代码只求出sa[ ], rank[ ], height[ ]#include #include const int maxn=200;char s[maxn];//假设字符串中出现的都是大写英文字母int sa[maxn],rank[maxn],height[maxn],wa[maxn],wb[maxn],wv[maxn],ws[maxn],n;int cmp(int *r,int

思路：二分子串的长度。判断L长度的子串是否出现k的办法是对height数组进行分段，分段的依据是height[ i ]>=L,若存在某段中元素的个数>=k-1,则存在长度为L的子串在主串中至少出现k次。#include#include#include#includeusing namespace std;const int maxn=20005;int s[maxn],rank[m

#include#include#includeconst int maxn=30000005;typedef __int64 ll;int n,a[maxn];ll c[10005];int lowbit(int x){ return x & (-x);}void Update(int x){ while(x<=10000) { c[x]+

#include#include#includeconst int maxn=100005;typedef __int64 ll;struct node{ ll s,num;};node tnode[maxn];int n;int lowbit(int x){ return x&(-x);}node find(int x){ node ans;

#include#include#includeusing namespace std;const int maxn=100005;int a[maxn],c[maxn],mini1[maxn],mini2[maxn],n,m;int lowbit(int x){ return x&(-x);}int getsum(int x){ int ans=0; while(x>0

#include#include#include#define MIN(a,b) a>b?b:aconst int maxn=200005;int h,w,n;struct node{ int l,r,max;};node tnode[maxn*4];void Build(int o,int x,int y){ tnode[o].l=x; tnode[

#include#include#includeconst int maxn=50005;int a[maxn],c[maxn],n;int lowbit(int x){ return x&(-x);}int find(int x){ if(x>0) { return c[x]+find(x-lowbit(x)); } re

#include#includeconst int maxn = 1005;int c[maxn][maxn],n,t,T;int lowbit(int x){ return x & (-x);}void update(int x,int y,int d){ if(x==0 || y==0) return; for(int i=x;i<=n;i+=lowbit(i)) {