Color the ball
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18915 Accepted Submission(s): 9442
當N = 0,輸入結束。
二段樹,但是我不會,有這種區間的簡單做法
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
int main(){
int n,i,sum[100010],a,b;
while(scanf("%d",&n)&&n!=0){
memset(sum,0,sizeof(sum));
for(i=1;i<=n;i++){
scanf("%d%d",&a,&b);
sum[a]++;
sum[++b]--;
}
for(i=1;i<=n;i++){
sum[i]+=sum[i-1];
if(i==1) printf("%d",sum[i]);
else printf(" %d",sum[i]);
}
printf("\n");
}
return 0;
}