hdoj--1556--Color the ball(模拟&&树状数组)

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Color the ball

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13727 Accepted Submission(s): 6903

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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先来一组大神的代码，真神奇的模拟，思路太好了

#include<stdio.h>#include<string.h>int num[1000010];int main(){int n;while(scanf("%d",&n),n){memset(num,0,sizeof(num));int m=0;for(int i=0;i<n;i++){int a,b;scanf("%d%d",&a,&b);num[a]++;num[b+1]--;}for(int i=1;i<n;i++){m+=num[i];printf("%d ",m);}printf("%d\n",m+num[n]);}return 0;}

树状数组

#include<stdio.h>#include<string.h>int num[100010];int n;void add(int x,int k){while(x>=1){num[x]+=k;x-=x&(-x);}}int sum(int x){int s=0;while(x<=n){s+=num[x];x+=x&(-x);}return s;}int main(){while(scanf("%d",&n),n){memset(num,0,sizeof(num));for(int i=0;i<n;i++){int a,b;scanf("%d%d",&a,&b);add(b,1);add(a-1,-1);}for(int i=1;i<=n;i++){if(i>1)printf(" ");printf("%d",sum(i));}printf("\n");}return 0;}

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