UVA 12003 Array Transformer

Array Transformer

Time Limit: 5000ms

Memory Limit: 131072KB

This problem will be judged on UVA. Original ID: 12003
64-bit integer IO format: %lld      Java class name: Main

 

Write a program to transform an array A[1], A[2],..., A[n] according to m instructions. Each instruction (L, R, v, p) means: First, calculate how many numbers from A[L] to A[R](inclusive) are strictly less than v, call this answer k. Then, change the value of A[p] to u*k/(R - L + 1), here we use integer division (i.e. ignoring fractional part).

Input

The first line of input contains three integer n, m, u ( 1n300, 000, 1m50, 000, 1u1, 000, 000, 000). Each of the next n lines contains an integer A[i] ( 1A[i]u). Each of the next m lines contains an instruction consisting of four integers L, R, v, p ( 1LRn, 1vu, 1pn).

Output

Print n lines, one for each integer, the final array.

Sample Input

10 1 11
1
2
3
4
5
6
7
8
9
10
2 8 6 10

Sample Output

1
2
3
4
5
6
7
8
9
6

Explanation: There is only one instruction: L = 2, R = 8, v = 6, p = 10. There are 4 numbers (2,3,4,5) less than 6, so k = 4. The new number in A[10] is 11*4/(8 - 2 + 1) = 44/7 = 6.

解题：分块

参考lrj同学的那本大白

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int maxn =  + ;
 const int SIZE = ;
 int n,m,u,A[maxn],block[maxn/SIZE+][SIZE];
 void init() {
     scanf("%d%d%d",&n,&m,&u);
     int b = ,j = ;
     for(int i = ; i < n; ++i) {
         scanf("%d",A+i);
         block[b][j] = A[i];
         if(++j == SIZE) {
             b++;
             j = ;
         }
     }
     for(int i = ; i < b; ++i)
         sort(block[i],block[i] + SIZE);
     if(j) sort(block[b],block[b]+j);
 }
 int query(int L,int R,int v) {
     int lb = L/SIZE,rb = R/SIZE,k = ;
     if(lb == rb) {
         for(int i = L; i <= R; ++i)
             k += (A[i] < v);
     } else {
         for(int i = L; i < (lb+)*SIZE; ++i)
             if(A[i] < v) ++k;
         for(int i = rb*SIZE; i <= R; ++i)
             if(A[i] < v) ++k;
         for(int i = lb+; i < rb; ++i)
             k += lower_bound(block[i],block[i]+SIZE,v) - block[i];
     }
     return k;
 }
 void update(int p,int x) {
     if(A[p] == x) return;
     int old = A[p],pos = ,*B = &block[p/SIZE][];
     A[p] = x;
     while(B[pos] < old) ++pos;
     B[pos] = x;
     while(pos < SIZE- && B[pos] > B[pos + ]) {
         swap(B[pos],B[pos+]);
         ++pos;
     }
     while(pos >  && B[pos] < B[pos - ]) {
         swap(B[pos],B[pos-]);
         --pos;
     }
 }
 int main() {
     init();
     while(m--) {
         int L,R,v,p;
         scanf("%d%d%d%d",&L,&R,&v,&p);
         --L;
         --R;
         --p;
         int k = query(L,R,v);
         update(p,(LL)u*k/(R - L + ));
     }
     for(int i = ; i < n; ++i)
         printf("%d\n",A[i]);
     return ;
 }