#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 100001
#define L 31
#define INF 1000000009
#define eps 0.00000001
struct node
{
LL l, r, data;
}T[MAXN*+];
LL n,a[MAXN];
LL Query(LL p, LL k)
{
if (T[p].l == T[p].r)
return T[p].data;
LL mid = (T[p].l + T[p].r) >> ;
LL sum = T[p].data;
if (k <= mid)
sum += Query(p << , k);
else
sum += Query(p << | , k);
return sum;
}
void Build(LL p, LL l, LL r)
{
T[p].l = l, T[p].r = r, T[p].data = ;
if (l == r)
{
T[p].data = a[l];
return;
}
LL mid = (l + r) >> ;
Build(p << , l, mid);
Build(p << | , mid + , r);
}
void Insert(LL p, LL l, LL r, LL num)
{
//cout << p << ' ' << l << ' ' << r << ' ' << num << endl;
if (l <= T[p].l&&r >= T[p].r)
{
T[p].data += num;
return;
}
LL mid = (T[p].l + T[p].r) / ;
if (r <= mid)
Insert(p << , l, r, num);
else if (l > mid)
Insert(p << | , l, r, num);
else
{
Insert(p << , l, mid, num);
Insert(p << | , mid + , r, num);
}
}
int main()
{
while (scanf("%lld", &n), n)
{
LL t1, t2;
memset(a, , sizeof(a));
Build(, , n);
for (LL i = ; i <= n; i++)
{
scanf("%lld%lld", &t1, &t2);
Insert(, t1, t2, );
}
for (LL i = ; i <= n; i++)
{
if (i>) printf(" ");
printf("%lld", Query(, i));
}
printf("\n");
}
return ;
}
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Color the ballTime Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) 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 Author
8600
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