HDU 1556 Color the ball-from lanshui_Yang

Problem Description
N balloons are arranged in a row, numbered 1, 2, 3... from left to right .... n. given two integers a B (a <= B) each time, lele colors each balloon one time from balloon a to balloon B, riding his "little pigeon" electric car. But after N times, lele has forgotten how many times the I-th balloon has been painted. Can you help him figure out how many times each balloon has been painted?

Input
The first behavior of each test instance is an integer N (N <= 100000 ). next N rows, each row contains two integers, a B (1 <= a <= B <= N ).
When N = 0, the input ends.

Output
Each test instance outputs a row, which contains N integers. The I number indicates the total number of times that the I balloon is colored.

Sample Input
3
1 1
2 2
3 3
3
1 1
1 2
1 3
0

Sample Output
1 1 1
3 2 1

This is a typical one-dimensional tree array deformation. The common one-dimensional tree array is used for single-point update and interval evaluation. This question is another use of the tree array: Interval Update, single-point evaluation. The principle is as follows:

Assume that the elements of the original array are a [1], a [2],… A [n], then d [n] = a [1] + a [2] + ...... + A [n] calculates the sum of the First n items. This is the first purpose of the tree array: single-point update, interval summation.

Then, make some changes. Assume that the elements of the original array are a [1]-0, a [2]-a [1], a [3]-a [2], ......, A [n]-a [n-1], then the first n items and d [n] = a [n], that is, now the first n items and d [n] of the original array are equal to the single point value a [n]. do you understand this?

Next, If You Want To [a [m]... If all values in a [n] + Val, you only need to add Val to the m entry (a [m]-a [m-1]) of the original array, and subtract Val from n + 1 (a [n + 1]-a [n]), so that when m <= I <= n,

The first I and:

D [I] = (a [1]-0) + (a [2]-a [1]) + (a [3]-a [2]) + ...... + (A [m]-a [m-1] + val) + (a [m + 1]-a [m]) + ...... + (A [I]-a [I-1]) = a [I] + val.

Similarly, when I> n, d [I] is equal to the original a [I]. If you can see this, you will be very open. Note that a [1]... The initial values of a [n] are 0 !!

See the code below:

 

#include<iostream>#include<cstring>#include<string>#include<cstdio>#include<cmath>#include<algorithm>#include<queue>using namespace std ;const int MAXN = 1e5 + 5 ;int C[MAXN] ;int n ;int lowbit (int x){    return x & -x ;}void add(int x , int d){    while(x <= n)    {        C[x] += d ;        x += lowbit(x) ;    }}int sum(int x){    int sumt = 0 ;    while (x > 0)    {        sumt += C[x] ;        x -= lowbit(x) ;    }    return sumt ;}int main(){    while (scanf("%d" , &n) != EOF)    {        if(n == 0 )            break ;        memset(C , 0 , sizeof(C)) ;        int t = n ;        int i ;        while ( t-- )        {            int a , b ;            scanf("%d%d", &a , &b) ;            add(a , + 1) ;            add(b + 1 , -1) ;        }        for(i = 1 ; i <= n ; i ++)        {            printf("%d" , sum(i)) ;            if(i < n)                printf(" ") ;        }        puts("") ;    }    return 0 ;}