摘要：#include <iostream>#include <algorithm>#include <stdio.h>#include <string>#include <string.h>#include <stdlib.h>#include <math.h>#include <vector>#include <map>#include <queue>#include <stack>#include <set>using namespace std ;c 阅读全文

摘要：#include <iostream>#include <stdio.h>#include <string>#include <string.h>#include <algorithm>#include <math.h>#include <fstream>#include <vector>#include <map>#include <queue>#include <stack>#include <math.h>#include <stdlib. 阅读全文

摘要：这道题拿过来感觉很蒙 ， 因为以前做的都是点更新询问区间 ， 或是区间更新询问点 ， 这道题是在[a,b]区间内隔k个数更新一次（(i - a) % k == 0即i%k==a%k）,对于树状数组来说按照一维的方式定义， 后面在加上两维 c[i][k][i%k] (a<=i<=b)， 相当于在一维c[i]每个节点加入了一些信息，来记录每次更新，询问时候再加上原数组即可得到结果。#include <iostream>#include <stdio.h>#include <string>#include <string.h>#includ 阅读全文

摘要：二分+树状数组动态的寻找比a大k的数#include <stdio.h>#include <iostream>#include <string>#include <string.h>#include <algorithm>#include <stdlib.h>#include <math.h>#include <vector>#include <map>using namespace std;const int maxn = 100005;int c[maxn];int lowbit(i 阅读全文

摘要：/* s1[i]与s2[i]匹配，树状数组i位置更新1，否则更新0。*/#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;const int maxn = 1000005;int c[maxn];char s1[maxn],s2[maxn];int min(int a,int b){ return a>b?b:a;}int lowbit(int x){ return x&(-x);}void upda 阅读全文