大白上的原题,我就练练手。。。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 3e5 + 10;
const int SIZE = 4096;
ll block[N / SIZE + 1][SIZE + 1];
ll A[N];
int query(int L, int R, int v)
{
int k = 0;
int lb = L / SIZE, rb = R / SIZE;
if(lb == rb) { for(int i = L; i <= R; ++i) if(A[i] < v) k++; }
else {
for(int i = L; i < (lb + 1) * SIZE; ++i) if(A[i] < v) k++;
for(int i = rb * SIZE; i <= R; ++i) if(A[i] < v) k++;
for(int i = lb + 1; i < rb; ++i) {
k += lower_bound(block[i], block[i] + SIZE, v) - block[i];
}
}
return k;
}
void change(int k, int u, int p, int L, int R)
{
ll x = (ll)u * k / (R - L + 1);
if(A[p] == x) return;
int la = p / SIZE;
ll* B = &block[la][0];
int pos = 0;
ll old = A[p];
while(B[pos] < old) pos++;
A[p] = x; B[pos] = x;
if(x > old) {
while(pos < SIZE - 1 && B[pos] > B[pos + 1]) { swap(B[pos + 1], B[pos]); pos++; }
}
else {
while(pos > 0 && B[pos] < B[pos - 1]) { swap(B[pos - 1], B[pos]); pos--; }
}
}
int main()
{
int n, m, u;
while(~scanf("%d%d%d", &n, &m, &u))
{
int j = 0, k = 0;
for(int i = 0; i < n; ++i)
{
scanf("%lld", &A[i]);
block[k][j++] = A[i];
if(j == SIZE) { k++; j = 0; }
}
for(int i = 0; i < k; ++i) sort(block[i], block[i] + SIZE);
if(j) sort(block[k], block[k] + j);
int L, R, v, p;
while(m --)
{
int ans = 0;
scanf("%d%d%d%d", &L, &R, &v, &p);
L--; R--; p--;
ans = query(L, R, v);
change(ans, u, p, L, R);
}
for(int i = 0; i < n; ++i) printf("%lld\n", A[i]);
}
return 0;
}
原文:http://www.cnblogs.com/orchidzjl/p/4905541.html