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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.
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

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：由于數最大不超過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; }

　　

轉載于:https://www.cnblogs.com/ZhaoxiCheung/p/7352039.html

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