Color the ball

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19849 Accepted Submission(s): 9897

Problem Description

N个气球排成一排，从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球a开始到气球b依次给每个气球涂一次颜色。但是N次以后lele已经忘记了第I个气球已经涂过几次颜色了，你能帮他算出每个气球被涂过几次颜色吗？

Input

每个测试实例第一行为一个整数N,(N <= 100000).接下来的N行，每行包括2个整数a b(1 <= a <= b <= N)。
当N = 0，输入结束。

Output

每个测试实例输出一行，包括N个整数，第I个数代表第I个气球总共被涂色的次数。

Sample Input

31 12 23 331 11 21 30

Sample Output

1 1 13 2 1

Author

8600

Source

HDU 2006-12 Programming Contest

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代码写出来了，但是一开始不知道原理是什么，然后突然看到了另一份代码就懂了。

#include<stdio.h>  #include<string.h>  int s[100100];  int main(){      int n;      while(~scanf("%d",&n)&&n){          int i,j,k;          int x,y;          memset(s,0,sizeof(s));          for(i=1;i<=n;i++){              scanf("%d %d",&x,&y);              s[x]++;              s[y+1]--;          }          int sum=0;          for(i=1;i<n;i++){              sum+=s[i];              printf("%d ",sum);          }          sum+=s[i];          printf("%d\n",sum);      }      return 0;  } 

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;const int Maxn=100000;int sum[1000005],n;int lowbit(int x){    return x&(-x);}void updata(int x,int cnt){    while(x<=Maxn){        sum[x]+=cnt;        x+=lowbit(x);    }}int getSum(int x){    int s=0;    while(x>0){        s+=sum[x];        x-=lowbit(x);    }    return s;}int main(){   // freopen("in.txt","r",stdin);    while(scanf("%d",&n)&&n){        memset(sum,0,sizeof(sum));        for(int l,r,i=1;i<=n;i++){            scanf("%d%d",&l,&r);            updata(l,1);            updata(r+1,-1);        }        for(int i=1;i<n;i++) printf("%d ",getSum(i));        printf("%d\n",getSum(n));    }}