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二维树状数组：同样不要忘记c的初始化，modify的功能是改变元素(x,y)，sum的功能则是求从元素(1,1)开始到(x,y)的总和，同样，可以求出任意一个子矩阵内的所有元素之和，即sum(x2,y2)-sum(x1-1,y2)-sum(x2,y1-1)+sum(x1-1,y1-1)intlowbit(intx){returnx&(-x);}voidmodify(intx,inty,intdelta){inti,j;for(i=x;i<=N;i+=lowbit(i)){for(j=y;j<=N;j+=lowbit(j)){c[i][j]+=delta;}}}intsum( 阅读全文
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http://acm.hdu.edu.cn/showproblem.php?pid=1541题意：给定一些点，求在这点左下方的点的个数;思路：此题可以用二维树状数组做，但仔细看题，可以得到给出的y是从小到大的，这样，只需对x进行操作，进而转换为在x左边的点有多少个代码：#include<iostream>using namespace std;int a[32005],c[32005],n=32005;int lowbit(int x){ return x&(-x);}int sum(int x){ int sum=0; while(x>0) { sum+=c[x]; 阅读全文
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http://acm.hdu.edu.cn/showproblem.php?pid=1394题意：求最小逆序数#include<iostream>using namespace std;int c[5005],n,a[5005];int lowbit(int x){ return x&(-x);}int sum(int x){ int sum=0; while(x>0) { sum+=c[x]; x-=lowbit(x); } return sum;}void inster(int x,int i){ while(x<=n) { c[x]+=i; x+=lowb 阅读全文
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http://acm.hdu.edu.cn/showproblem.php?pid=2838思路：对于数据3，4，1，2先按顺序建树，updata(a[i],a[i]),s+=a[i];当i=3时，1前面有3，4比1大，这时，先求出比1等于或小的元素和，在用s-sum(1),结果为，1前面比1大的元素和。代码：View Code 1 #include<iostream> 2 using namespace std; 3 __int64 a[100005],c[100005],d[100005],n; 4 int lowbit(int x) 5 { 6 return x&(- 阅读全文
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http://acm.hdu.edu.cn/showproblem.php?pid=2689代码：#include<iostream>using namespace std;int c[1010],a[1010],n;int lowbit(int x){ return x&(-x);}int sum(int x){ int sum=0; while(x>0) { sum=sum+c[x]; x-=lowbit(x); } return sum;}void inster(int x,int i){ while(x<=n) { c[x]+=i; x+=lowbit( 阅读全文
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http://acm.hdu.edu.cn/showproblem.php?pid=1556代码：#include<iostream>int n,c[100003];int lowbit(int x){ return x&(-x);}int sum(int x){ int sum=0; while(x>0) { sum+=c[x]; x-=lowbit(x); } return sum;}void inster(int x,int i){ while(x<=n) { c[x]+=i; x+=lowbit(x); }}int main(){ int a,b,i; 阅读全文
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http://acm.hdu.edu.cn/showproblem.php?pid=1166代码：#include<iostream>using namespace std;int n,a[50005],c[50005];int lowbit(int x){ return x&(-x);}int sum(int x){ int sum=0; while(x>0) { sum+=c[x]; x=x-lowbit(x); } return sum;}void inster(int i,int j){ while(i<=n) ... 阅读全文
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Ping pongTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2242Accepted Submission(s): 823Problem DescriptionN(3<=N<=20000) ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill r 阅读全文