LOJ#6281. 数列分块入门 5
数据最大就2^31次方,通过计算这些数开方五次就变成1了,然后把继续开方也不会变了,所以对于每个块,前5次更新暴力,之后就可以直接跳过,查询就直接就可以了
#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 = 50005; const int maxm = 100000; const int mod = 998244353; using namespace std; int n, m, tol, T; int block; int a[maxn]; int b[maxn]; int c[maxn]; int belong[maxn]; void init() { memset(a, 0, sizeof a); memset(b, 0, sizeof b); memset(c, 0, sizeof c); memset(belong, 0, sizeof belong); } int L(int x) { return (x-1) * block + 1; } int R(int x) { return min(n, x*block); } void update(int l, int r) { for(int i=l; i<=min(r, R(belong[l])); i++) { b[belong[i]] -= a[i]; a[i] = sqrt(a[i]); b[belong[i]] += a[i]; } if(belong[l] == belong[r]) return ; if(belong[l] != belong[r]) { for(int i=L(belong[r]); i<=r; i++) { b[belong[i]] -= a[i]; a[i] = sqrt(a[i]); b[belong[i]] += a[i]; } } for(int i=belong[l]+1; i<belong[r]; i++) { if(c[i] == 6) continue; for(int j=L(i); j<=R(i); j++) { b[i] -= a[j]; a[j] = sqrt(a[j]); b[i] += a[j]; } c[i]++ ; } } int query(int l, int r) { int ans = 0; for(int i=l; i<=min(r, R(belong[l])); i++) ans += a[i]; if(belong[l] == belong[r]) return ans; if(belong[l] != belong[r]) { for(int i=L(belong[r]); i<=r; i++) { ans += a[i]; } } for(int i=belong[l]+1; i<belong[r]; i++) ans += b[i]; return ans; } int main() { while(~scanf("%d", &n)) { block = sqrt(n); for(int i=1; i<=n; i++) { scanf("%d", &a[i]); belong[i] = (i-1) / block + 1; b[belong[i]] += a[i]; } m = n; while(m--) { int op, l, r, c; scanf("%d%d%d%d", &op, &l, &r, &c); if(op == 0) { update(l, r); } else { int ans = query(l, r); printf("%d\n", ans); } } } return 0; }