HDU1556 Color the ball(树状数组+抖机灵)
2017-04-21 17:52
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题目:
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19338 Accepted Submission(s): 9651
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
Source
HDU 2006-12 Programming Contest
Recommend
LL | We have carefully selected several simi
f6ef
lar problems for you: 1542 1394 1698 1255 2795
Statistic | Submit | Discuss | Note
题目:
简单的树状数组的题,但是用普通的方法竟然能过,不知道为啥
代码1(树状数组):
#include <cstdio>
#include <cstring>
#include <cctype>
#include <string>
#include <set>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define mod 10000007
#define debug() puts("what the fuck!!!")
#define N 100000+20
#define M 1000000+10
#define ll long long
using namespace std;
int c
;
int lowbit(int x)//用来维护树状数组,得到管辖区间
{
return x&-x;
}
void add(int k,int num,int n)//给数组的第k位增加num个
{
while(k<=n)
{
c[k]+=num;
k+=lowbit(k);
}
}
int sum(int n)//求要查询的区间的前n项和
{
int sum=0;
while(n>0)
{
sum+=c
;
n-=lowbit(n);
}
return sum;
}
int main()
{
int n,a,b;
while(scanf("%d",&n)&&n)
{
mem(c,0);
for(int i=1; i<=n; i++)
{
scanf("%d%d",&a,&b);
add(a,1,n);
add(b+1,-1,n);//主要是这里,向上统计,向下修改
}
for(int i=1; i<=n; i++)
{
if(i>1)printf(" ");
printf("%d",sum(i));
}
puts("");
}
return 0;
}代码2(普通的数组):
#include <cstdio>
#include <cstring>
#include <cctype>
#include <string>
#include <set>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define mod 10000007
#define debug() puts("what the fuck!!!")
#define N 100000+20
#define M 1000000+10
#define ll long long
using namespace std;
int c
;
int main()
{
int n,a,b;
while(scanf("%d",&n)&&n)
{
mem(c,0);
for(int i=1; i<=n; i++)
{
scanf("%d%d",&a,&b);
c[a]++;
c[b+1]--;
}
int ans=0;
for(int i=1; i<=n; i++)
{
if(i>1)printf(" ");
ans+=c[i];
printf("%d",ans);
}
puts("");
}
return 0;
}
这两个为什么后面的用的时间还比前面的要少?不是树状数组会更快么?
Color the ball
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 19338 Accepted Submission(s): 9651
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
Source
HDU 2006-12 Programming Contest
Recommend
LL | We have carefully selected several simi
f6ef
lar problems for you: 1542 1394 1698 1255 2795
Statistic | Submit | Discuss | Note
题目:
简单的树状数组的题,但是用普通的方法竟然能过,不知道为啥
代码1(树状数组):
#include <cstdio>
#include <cstring>
#include <cctype>
#include <string>
#include <set>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define mod 10000007
#define debug() puts("what the fuck!!!")
#define N 100000+20
#define M 1000000+10
#define ll long long
using namespace std;
int c
;
int lowbit(int x)//用来维护树状数组,得到管辖区间
{
return x&-x;
}
void add(int k,int num,int n)//给数组的第k位增加num个
{
while(k<=n)
{
c[k]+=num;
k+=lowbit(k);
}
}
int sum(int n)//求要查询的区间的前n项和
{
int sum=0;
while(n>0)
{
sum+=c
;
n-=lowbit(n);
}
return sum;
}
int main()
{
int n,a,b;
while(scanf("%d",&n)&&n)
{
mem(c,0);
for(int i=1; i<=n; i++)
{
scanf("%d%d",&a,&b);
add(a,1,n);
add(b+1,-1,n);//主要是这里,向上统计,向下修改
}
for(int i=1; i<=n; i++)
{
if(i>1)printf(" ");
printf("%d",sum(i));
}
puts("");
}
return 0;
}代码2(普通的数组):
#include <cstdio>
#include <cstring>
#include <cctype>
#include <string>
#include <set>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define mod 10000007
#define debug() puts("what the fuck!!!")
#define N 100000+20
#define M 1000000+10
#define ll long long
using namespace std;
int c
;
int main()
{
int n,a,b;
while(scanf("%d",&n)&&n)
{
mem(c,0);
for(int i=1; i<=n; i++)
{
scanf("%d%d",&a,&b);
c[a]++;
c[b+1]--;
}
int ans=0;
for(int i=1; i<=n; i++)
{
if(i>1)printf(" ");
ans+=c[i];
printf("%d",ans);
}
puts("");
}
return 0;
}
这两个为什么后面的用的时间还比前面的要少?不是树状数组会更快么?
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