hdu1166题解

这道入门题十分经典，基本给出线段树的模板了（建树，改变树，取值）
很容易看出这道题区间维护的是和。

#include 
#include 
int tree[150001],N,res[40000];
void build(int tn,int left,int right)
{
    int mid;
    if(left==right) {   scanf("%d",tree+tn);return;}
    mid=(left+right)>>1;
    build(tn<<1,left,mid);
    build(tn<<1|1,mid+1,right);
    tree[tn]=tree[tn<<1]+tree[tn<<1|1];
}
void chang(int tn,int left,int right,int x,int num)
{
    int mid;
    if(left==right) {tree[tn]+=num;return;  }
    mid=(left+right)>>1;
    if(x<=mid) chang(tn<<1,left,mid,x,num);
    else chang(tn<<1|1,mid+1,right,x,num);
    tree[tn]=tree[tn<<1]+tree[tn<<1|1];//必须更新父节点
}
int find(int tn,int left,int right,int x,int y)
{
    int mid;
    if(x<=left&&y>=right) return tree[tn];
    mid=(left+right)>>1;
    if(y<=mid) return find(tn<<1,left,mid,x,y);
    else if(x>mid) return find(tn<<1|1,mid+1,right,x,y);
    else return find(tn<<1,left,mid,x,y)+find(tn<<1|1,mid+1,right,x,y);
}
int main(int argc, char const *argv[])
{
    int i=0,j,T,a,b,k;
    char str[10];
    scanf("%d",&T);
    while(T--)
{   printf("Case %d:\n",++i );
    k=0;
    memset(tree,0,150001*sizeof(int));
    scanf("%d",&N);
    build(1,1,N);
    while(scanf("%s",str)!=EOF&&strcmp(str,"End"))
    {
        scanf("%d %d",&a,&b);
        if(strcmp(str,"Add")==0)
            chang(1,1,N,a,b);
        else if(strcmp(str,"Sub")==0)
            chang(1,1,N,a,-b);
        else res[k++]=find(1,1,N,a,b);
    }
    for(j=0;jprintf("%d\n",res[j] );
}
    return 0;
}