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
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。
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
3 1 1 2 2 3 3 3 1 1 1 2 1 3 0
Sample Output
1 1 1 3 2 1
Author
8600
Source
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先来一组大神的代码,真神奇的模拟,思路太好了
树状数组
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 ",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(" "); } return 0; }