线段树--区间更新区间查询--hdu4027

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题意：

给定一串数字,再给定两个操作：

0.查询 x 到 y的和

1.更新 x到 y的每个值，使其变为根号倍

Notice that the square root operation should be rounded down to integer.

将有可能变成的小树变为整数

#include <iostream>#include <stdio.h>#include <math.h>using namespace std;#define lson i<<1,l,m#define rson i<<1|1,m+1,rlong long a[100010];long long ans;struct TreeNode{    long long  num;    int left;    int right;    int ac;}Tree[100010*4];void Build(int i, int l , int r){    Tree[i].left=l;    Tree[i].right=r;    if(l==r)    {       Tree[i].num=a[l];       if(Tree[i].num==1)        Tree[i].ac=1;        else        Tree[i].ac=0;       return ;    }    int m=(l+r)/2;    Build(lson);    Build(rson);    Tree[i].num=Tree[i<<1].num+Tree[i<<1|1].num;    if(Tree[i*2].ac&&Tree[i*2+1].ac)       Tree[i].ac=1;   else Tree[i].ac=0;}void U(int i,int l,int r){    if(Tree[i].ac)return;    if(Tree[i].left==Tree[i].right)    {        Tree[i].num=floor(sqrt(Tree[i].num));        if(Tree[i].num==1)            Tree[i].ac=1;        return ;    }    int m=(Tree[i].left+Tree[i].right)>>1;    if(l>m)    {        U(i<<1|1,l,r);    }    else if(r<=m)    {        U(i<<1,l,r);    }    else    {        U(lson);        U(rson);    }    Tree[i].num=Tree[i<<1].num+Tree[i<<1|1].num; if(Tree[i*2].ac&&Tree[i*2+1].ac)       Tree[i].ac=1;}void  Q(int i, int l ,int r){    if(l==Tree[i].left&&r==Tree[i].right)    {       ans+=Tree[i].num;        return ;    }    int m=(Tree[i].left+Tree[i].right)>>1;    if(r<=m)    {        Q(i<<1,l,r);    }    else if(l>m)    {        Q(i<<1|1,l,r);    }    else    {        Q(lson);        Q(rson);    }}int main(){    int n;    int flag=0;    while(scanf("%d",&n)!=EOF)    {   flag++;        printf("Case #%d:\n",flag);        int i;        for(i=1;i<=n;i++)        {            scanf("%lld",&a[i]);        }        Build(1,1,n);        int t;        scanf("%d",&t);        while(t--)        {        ans=0;        int d,x,y;        scanf("%d%d%d",&d,&x,&y);        if(x>y)            swap(x,y);        if(d==0)        {            U(1,x,y);        }        else        {Q(1,x,y);            printf("%lld\n",ans);        }        }        printf("\n");    }   return 0;}

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