题意：给定一个序列，然后有 m 个修改，问你最后的序列是什么，修改是这样的 l r v p 先算出从 l 到 r 这个区间内的 小于 v 的个数k，然后把第 p 个的值改成 k * u / (r - l + 1)。
析：分块，每块长度是sz，把每一块都排序。然后在每次修改的时候，只要计算出 l 和 r 所在块，中间的用二分可以算出来。注意同时要把分块中的数也改掉。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
//#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 3e5 + 10;
const int maxm = 1e5 + 10;
const int mod = 50007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn];
int b[560][560];
const int sz = 548;
int main(){
int u;
while(scanf("%d %d %d", &n, &m, &u) == 3){
int idx = 0, cnt = 0;
for(int i = 0; i < n; ++i){
scanf("%d", a + i);
b[idx][cnt] = a[i];
if(++cnt == sz){ ++idx; cnt = 0; }
}
for(int i = 0; i < idx; ++i) sort(b[i], b[i] + sz);
if(cnt) sort(b[idx], b[idx] + cnt);
while(m--){
int l, r, v, p;
scanf("%d %d %d %d", &l, &r, &v, &p);
--l, --r, --p;
int lx = l / sz, rx = r / sz;
int k = 0;
if(lx == rx){
for(int i = l; i <= r; ++i) if(v > a[i]) ++k;
}
else {
for(int i = l; i < (lx+1)*sz; ++i) if(v > a[i]) ++k;
for(int i = rx*sz; i <= r; ++i) if(v > a[i]) ++k;
for(int i = lx+1; i < rx; ++i)
k += lower_bound(b[i], b[i] + sz, v) - b[i];
}
int c = (LL)u * k / (r - l + 1);
int old = a[p];
a[p] = c;
int mx = p / sz;
p = 0;
while(b[mx][p] < old) ++p;
b[mx][p] = c;
while(p - 1 >= 0 && b[mx][p] < b[mx][p-1]) swap(b[mx][p], b[mx][p-1]), --p;
while(p + 1 < sz && b[mx][p] > b[mx][p+1]) swap(b[mx][p], b[mx][p+1]), ++p;
}
for(int i = 0; i < n; ++i) printf("%d
", a[i]);
}
return 0;
}