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Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 18290 Accepted Submission(s): 4308

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.

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)

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

10

1 2 3 4 5 6 7 8 9 10

5

0 1 10

1 1 10

1 1 5

0 5 8

1 4 8

Sample Output

Case #1:

19

7

6

思路

題意：有N個數，有兩種操作，當T = 0時，區間L到R的數都變為自己的開方（向下取整），當 T = 1時，計算出區間L到R的和。

思路：由於數最大不超過2的63次方，因此開六次就為1，因此通過lazy數組記錄當前的數能否繼續開放下去，不能的話，就不必更新，因為一直是1不變。

#include<bits/stdc++.h>
using namespace std;
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
typedef __int64 LL;
const int maxn = 100005;
LL sum[maxn<<2];
bool lazy[maxn<<2];

void PushUp(int rt){
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    lazy[rt] = lazy[rt << 1] && lazy[rt << 1 | 1];
}

void build(int l,int r,int rt){
    if (l == r){
        scanf("%I64d",&sum[rt]);
        return;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUp(rt);
}

void upd(int L,int R,int l,int r,int rt){
    if (l == r){
        sum[rt] = sqrt(sum[rt]);
        if (sum[rt] <= 1)   lazy[rt] = 1;
        return;
    }
    int m = (l + r) >> 1;
    if (L <= m && !lazy[rt << 1]) upd(L,R,lson);
    if (R > m && !lazy[rt << 1 | 1])  upd(L,R,rson);
    PushUp(rt);
}

LL qry(int L,int R,int l,int r,int rt){
    if (L <= l && r <= R){
        return sum[rt];
    }
    LL ret = 0;
    int m = (l + r) >> 1;
    if (L <= m) ret += qry(L,R,lson);
    if (R > m)  ret += qry(L,R,rson);
    return ret;
}

int main(){
    int cases = 1;
    int N,M,T,a,b;
    while (~scanf("%d",&N)){
        memset(lazy,false,sizeof(lazy));
        memset(sum,0,sizeof(sum));
        printf("Case #%d:\n",cases++);
        build(1,N,1);
        scanf("%d",&M);
        while (M--){
            scanf("%d%d%d",&T,&a,&b);
            if (a > b)  swap(a,b);
            if (T == 0){
                upd(a,b,1,N,1);
            }
            else{
                printf("%I64d\n",qry(a,b,1,N,1));
            }
        }
        printf("\n");
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
typedef __int64 LL;
const int maxn = 100005;
LL sum[maxn];

void PushUp(int rt){
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}

void build(int l,int r,int rt){
    if (l == r){
        scanf("%I64d",&sum[rt]);
        return;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUp(rt);
}

void upd(int L,int R,int l,int r,int rt){
    if (sum[rt] == r - l + 1)   return;
    if (l == r){
        sum[rt] = sqrt(sum[rt]);
        return;
    }
    int m = (l + r) >> 1;
    if (L <= m) upd(L,R,lson);
    if (R > m)  upd(L,R,rson);
    PushUp(rt);
}

LL qry(int L,int R,int l,int r,int rt){
    if (L <= l && r <= R){
        return sum[rt];
    }
    LL ret = 0;
    int m = (l + r) >> 1;
    if (L <= m) ret += qry(L,R,lson);
    if (R > m)  ret += qry(L,R,rson);
    return ret;
}

int main(){
    int cases = 1;
    int N,M,T,a,b;
    while (~scanf("%d",&N)){
        printf("Case #%d:\n",cases++);
        build(1,N,1);
        scanf("%d",&M);
        while (M--){
            scanf("%d%d%d",&T,&a,&b);
            if (a > b)	swap(a,b);
            if (T == 0){
                upd(a,b,1,N,1);
            }
            else{
                printf("%I64d\n",qry(a,b,1,N,1));
            }
        }
    }
    return 0;
}

HDU 4027 Can you answer these queries?（線段樹/區間不等更新）