杭电4027 Can you answer these queries?（线段树）

Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 9170    Accepted Submission(s): 2096

Problem Description

A lot of battleships of evil are arranged in a line before the battle. Our commander decides to use our secret weapon to eliminate the battleships. Each of the battleships can be marked a value of endurance. For every attack of our secret weapon, it could decrease the endurance of a consecutive part of battleships by make their endurance to the square root of it original value of endurance. During the series of attack of our secret weapon, the commander wants to evaluate the effect of the weapon, so he asks you for help.
You are asked to answer the queries that the sum of the endurance of a consecutive part of the battleship line.

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

 

Input

The input contains several test cases, terminated by EOF.
  For each test case, the first line contains a single integer N, denoting there are N battleships of evil in a line. (1 <= N <= 100000)
  The second line contains N integers Ei, indicating the endurance value of each battleship from the beginning of the line to the end. You can assume that the sum of all endurance value is less than 2⁶³.
  The next line contains an integer M, denoting the number of actions and queries. (1 <= M <= 100000)
  For the following M lines, each line contains three integers T, X and Y. The T=0 denoting the action of the secret weapon, which will decrease the endurance value of the battleships between the X-th and Y-th battleship, inclusive. The T=1 denoting the query of the commander which ask for the sum of the endurance value of the battleship between X-th and Y-th, inclusive.

 

Output

For each test case, print the case number at the first line. Then print one line for each query. And remember follow a blank line after each test case.

 

Sample Input

101 2 3 4 5 6 7 8 9 1050 1 101 1 101 1 50 5 81 4 8

 

Sample Output

Case #1:1976

 

Source

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest 

/*题目不错，注意下更新到每个都是1的时候就没必要再往下访问开根号啦。此题坑爹。原因如下：1、输入左右大小不确定，线段树中返回的还是TLE2、尼玛，要求是每组案例后边输出一个空格，我在每次询问后输出空格，结果给我返回的是WA，而不是PE......Time:2015-1-20 9:17*///输入左右不一定是左小右大的结果是超时。。。。。无语。。#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1typedef long long LL;const int MAX=100000+10;struct Tree{    LL nSum;}tree[MAX<<2];void Build(int l,int r,int rt){    if(l==r){        scanf("%lld",&tree[rt].nSum);        return;    }    int mid=(l+r)>>1;    Build(lson);    Build(rson);    tree[rt].nSum=tree[rt<<1].nSum+tree[rt<<1|1].nSum;}void update(int L,int R,int l,int r,int rt){    if(l==L&&r==R&&tree[rt].nSum==r-l+1)return;    if(l==r){        tree[rt].nSum=sqrt(tree[rt].nSum*1.0);        return;    }    int mid=(l+r)>>1;    if(mid>=R)update(L,R,lson);    else if(mid<L)update(L,R,rson);    else{        update(L,mid,lson);        update(mid+1,R,rson);    }    tree[rt].nSum=tree[rt<<1].nSum+tree[rt<<1|1].nSum;}LL Query(int L,int R,int l,int r,int rt){    if(l==L&&R==r){        return tree[rt].nSum;    }    int mid=(l+r)>>1;    if(R<=mid) return Query(L,R,lson);    else if(L>mid) return Query(L,R,rson);    else return Query(L,mid,lson)+Query(mid+1,R,rson);}int main(){    int n,m;    int op,nCase=1;    while(scanf("%d",&n)!=EOF){        Build(1,n,1);        scanf("%d",&m);        int L,R;        printf("Case #%d:\n",nCase++);        while(m--){            scanf("%d%d%d",&op,&L,&R);            if(L>R)swap(L,R);            if(op){                printf("%lld\n",Query(L,R,1,n,1));            }else{                update(L,R,1,n,1);            }        }puts("");    }return 0;}