地址:http://acm.hdu.edu.cn/showproblem.php?pid=1166
树状数组基础题
include<stdio.h>
#include <iostream>
#include <stdio.h>
#include <cmath>
using namespace std;
int a[50045],node[50045];
int lowbit(int x) { return x & -x; }
void build(int a[], int node[], int size)
{
//将数组a处理成前缀和的形式
for (int i = 1; i <= size; i++)
a[i] += a[i - 1];
//根据每个节点覆盖的范围求出对应的值
for (int i = 1; i <= size; i++)
node[i] = a[i] - a[i - lowbit(i)];
}
void add(int pos, int val, int node[], int size)
{
//递推求出每个父节点,逐一更新直至达到根节点
for (int i = pos; i <= size; i += lowbit(i))
node[i] += val;
}
int prefix_sum(int n, int node[])
{
int sum = 0;
//从当前节点开始,每次向前求lowbit(i)个数之和,直至到达第一个节点
for (int i = n; i >= 1; i -= lowbit(i))
sum += node[i];
return sum;
}
int sum_between(int l, int r, int node[])
{
//则区间[l,r]的和可表示为前r项的和减去前l-1项和
return prefix_sum(r, node) - prefix_sum(l - 1, node);
}
int main()
{
int t,n1;
scanf("%d",&t);
for(n1=1;n1<=t;n1++)
{
int n2,n,i,j,x;
char s[10];
printf("Case %d:\n",n1);
scanf("%d",&n);
for(n2=1;n2<=n;n2++)
{
scanf("%d",&a[n2]);
}
build(a,node,n);
while(scanf("%s",s)!=EOF&&s[0]!='E')
{
scanf("%d%d",&i,&j);
if(s[0]=='A')
{
add(i,j,node,n);
}
if(s[0]=='S')
{
add(i,j*(-1),node,n);
}
if(s[0]=='Q')
{
x=sum_between(i,j,node);
printf("%d\n",x);
}
}
}
return 0;
}```