http://acm.hdu.edu.cn/showproblem.php?pid=1556
Color the ball
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6567 Accepted Submission(s): 3448
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
3
1 1
2 2
3 3
3
1 1
1 2
1 3
0
Sample Output
1 1 1
3 2 1
#include<stdio.h> #include<string.h> #define maxn 500004 struct node { int left,right; int num; }; int Total[maxn]; node tree[3*maxn]; int out=0; void build(int left,int right,int i) { tree[i].left =left; tree[i].right =right; tree[i].num =0; if(tree[i].left ==tree[i].right ) return ; build(left,(left+right)/2,2*i); build((left+right)/2+1,right,2*i+1); } void UPDATA(int r,int i,int j) { int MID=(tree[r].left +tree[r].right )/2; if(tree[r].left==i&&tree[r].right ==j) tree[r].num++; else { if(j <= MID) UPDATA(2*r,i,j); else if( i > MID) UPDATA(2*r+1,i,j); else { UPDATA(2*r,i,MID); UPDATA(2*r+1,MID+1,j); } } } void SUM(int r) { int i; for(i =tree[r].left; i <= tree[r].right;i++) Total[i]+=tree[r].num; if(tree[r].left==tree[r].right) return; else { SUM(2*r); SUM(2*r+1); } } int main() { int n,r,l,i; while(~scanf("%d",&n)&&n!=0) { build(1,n,1); for(i=1;i<=n;i++) { scanf("%d%d",&r,&l); UPDATA(1,r,l); } memset(Total,0,sizeof(Total)); SUM(1); for(i = 1; i <= n; i++) { if(i==1) printf("%d",Total[i]); else printf(" %d",Total[i]); } printf(" "); } return 0; }