标签：mat lowbit 杭电 max turn fine char dfs view
让ci = ai / bi, 求sum(ci)的值，因为每次 ai 都是加一的，那么我可以用一颗线段树来维护每个 i 位置的 ai 距离达到 bi 还需要的数的最小值，更新是每次都减一，如果我某一个区间的最小值等于 0， 这就说明我这时候的ai已经满足了ai/bi==1的情况，那么对应的ci的位置就应该加一，所以我每次发现一个区间的最小值是0的话，我就 dfs 去找到为0的地方，把这里重新修改成bi，然后这个位置的答案+1就可以了
#include<map> #include<set> #include<ctime> #include<cmath> #include<stack> #include<queue> #include<string> #include<vector> #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define first fi #define second se #define lowbit(x) (x & (-x)) typedef unsigned long long int ull; typedef long long int ll; const double pi = 4.0*atan(1.0); const int inf = 0x3f3f3f3f; const int maxn = 100005; const int maxm = 100000; const int mod = 998244353; using namespace std; int n, m, tol, T; int a[maxn]; int b[maxn]; int c[maxn << 2]; int sum[maxn << 2]; int lazy[maxn << 2]; void init() { memset(a, 0, sizeof a); memset(b, 0, sizeof b); memset(c, 0, sizeof c); memset(sum, inf, sizeof sum); memset(lazy, 0, sizeof lazy); } void pushup(int root) { sum[root] = min(sum[root << 1], sum[root << 1 | 1]); c[root] = c[root << 1] + c[root << 1 | 1]; } void pushdown(int root) { if(lazy[root] != 0) { lazy[root << 1] += lazy[root]; lazy[root << 1 | 1] += lazy[root]; sum[root << 1] -= lazy[root]; sum[root << 1 | 1] -= lazy[root]; lazy[root] = 0; } return ; } void build(int left, int right, int root) { if(left == right) { sum[root] = b[left]; return ; } int mid = (left + right) >> 1; build(left, mid, root << 1); build(mid+1, right, root << 1 | 1); pushup(root); } void dfs(int left, int right, int root) { if(left == right) { if(sum[root] == 0) { sum[root] = b[left]; c[root]++; } return ; } pushdown(root); int mid = (left + right) >> 1; if(sum[root << 1] == 0) dfs(left, mid, root << 1); if(sum[root << 1 | 1] == 0) dfs(mid+1, right, root << 1 | 1); pushup(root); } void update(int left, int right, int prel, int prer, int root) { if(prel <= left && right <= prer) { lazy[root]++; sum[root]--; while(sum[root] == 0) dfs(left, right, root); return ; } pushdown(root); int mid = (left + right) >> 1; if(prel <= mid) update(left, mid, prel, prer, root << 1); if(prer > mid) update(mid+1, right, prel, prer, root << 1 | 1); pushup(root); } int query(int left, int right, int prel, int prer, int root) { if(prel <= left && right <= prer) return c[root]; int mid = (left + right) >> 1; int ans = 0; if(prel <= mid) ans += query(left, mid, prel, prer, root << 1); if(prer > mid) ans += query(mid+1, right, prel, prer, root << 1 | 1); return ans; } int main() { while(~scanf("%d%d", &n, &m)) { init(); for(int i=1; i<=n; i++) scanf("%d", &b[i]); build(1, n, 1); char str[15]; while(m--) { int l, r; scanf("%s%d%d", str, &l, &r); if(str[0] == ‘a‘) { update(1, n, l, r, 1); } else if(str[0] == ‘q‘) { int ans = query(1, n, l, r, 1); printf("%d\n", ans); } } } return 0; }
Naive Operations HDU6315 （杭电多校2G）
标签：mat lowbit 杭电 max turn fine char dfs view
原文地址：https://www.cnblogs.com/H-Riven/p/9371067.html
