标签:name family pow str 个数 bsp sort ++ block
给你n个数,m次询问,Ks为区间内s的数目,求区间[L,R]之间所有Ks*Ks*s的和。1<=n,m<=200000.1<=s<=10^6
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 200050, INF = 0x7fffffff; int n, m, pos[maxn], s[maxn*10], c[maxn]; LL ans; struct node { int l, r, id; LL res; }Node[maxn]; bool cmp(node a, node b) { return pos[a.l] == pos[b.l] ? (a.r < b.r) : (a.l < b.l); } bool cmp_id(node a, node b) { return a.id < b.id; } void add(LL x) { s[c[x]]++; ans += c[x]*(s[c[x]]*s[c[x]] - (s[c[x]]-1)*(s[c[x]]-1)); } void del(LL x) { s[c[x]]--; ans -= c[x]*((s[c[x]]+1)*(s[c[x]]+1) - (s[c[x]])*(s[c[x]])); } int main() { ans = 0; scanf("%d%d", &n, &m); for(int i=1; i<=n; i++) scanf("%d", &c[i]); int block = sqrt(n); for(int i=1; i<=n; i++) pos[i] = (i-1)/block + 1; for(int i=1; i<=m; i++) { scanf("%d%d", &Node[i].l, &Node[i].r); Node[i].id = i; } sort(Node+1, Node+1+m, cmp); for(int i=1, l=1, r=0; i<=m; i++) { for(; r < Node[i].r; r++) add(r+1); for(; r > Node[i].r; r--) del(r); for(; l < Node[i].l; l++) del(l); for(; l > Node[i].l; l--) add(l-1); Node[i].res = ans; } sort(Node+1, Node+1+m, cmp_id); for(int i=1; i<=m; i++) printf("%I64d\n", Node[i].res); return 0; }
Powerful array CodeForces - 86D(莫队)
标签:name family pow str 个数 bsp sort ++ block
原文地址:https://www.cnblogs.com/WTSRUVF/p/9347681.html