hdu 4000 树状数组

Fruit Ninja

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 374    Accepted Submission(s): 197


[align=left]Problem Description[/align]
Recently, dobby is addicted in the Fruit Ninja. As you know, dobby is a free elf, so unlike other elves, he could do whatever he wants.But the hands of the elves are somehow strange, so when he cuts the fruit, he can only make specific move of his hands. Moreover, he can only start his hand in point A, and then move to point B,then move to point C,and he must make sure that point A is the lowest, point B is the highest, and point C is in the middle. Another elf, Kreacher, is not interested in cutting fruits, but he is very interested in numbers.Now, he wonders, give you a permutation of 1 to N, how many triples that makes such a relationship can you find ? That is , how many (x,y,z) can you find such that x < z < y ?
 
 

[align=left]Input[/align]
The first line contains a positive integer T(T <= 10), indicates the number of test cases.For each test case, the first line of input is a positive integer N(N <= 100,000), and the second line is a permutation of 1 to N.
 
 

[align=left]Output[/align]
For each test case, ouput the number of triples as the sample below, you just need to output the result mod 100000007.
 
 

[align=left]Sample Input[/align]

2 6 1 3 2 6 5 4 5 3 5 2 4 1

 
 

[align=left]Sample Output[/align]

Case #1: 10 Case #2: 1
判断满足i<j<k且num[i]<num[k]<num[j]的总组数
利用树状数组可以求出一个数前面比它小的数的个数，进而可以知道前面比它大的数的个数，总的比它大的个数减去前面比它大的个数等于后面比它大的个数，所以答案就很明显了~
 




View Code

#include<stdio.h>
#include<string.h>
int c[100010];
int n;
int lowbit(int x)
{
return x&(-x);
}
void update(int x,int d)
{
while(x<=n)
{
c[x]+=d;
x+=lowbit(x);
}
}
int sum(int x)
{
int ans=0;
while(x>0)
{
ans+=c[x];
x-=lowbit(x);
}
return ans;
}
int main()
{
__int64 ans;
int t,cases=1;
int i,a;
scanf("%d",&t);
while(t--)
{
ans=0;
memset(c,0,sizeof(c));
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%d",&a);
update(a,1);
int presmaller=sum(a-1);
int totbigger=n-a;
int prebigger=i-sum(a-1)-1;
__int64 afterbigger=totbigger-prebigger;
ans-=presmaller*afterbigger;
if(afterbigger>=2)
ans+=afterbigger*(afterbigger-1)/2;
}
printf("Case #%d: %I64d\n",cases++,ans%100000007);
}
return 0;
}