Can you answer these queries?Time

** limit:4000/2000 MS (java/others) Memory limit:65768/65768 K (java/others)**

Total submission (s): 10057 Accepted Submission (s): 2305

Problem Descriptiona lot of battleships of evil is arranged in a line before the battle. Our commander decides-to-use we secret weapon to eliminate the battleships. Each of the battleships can is marked a value of endurance. For every attack of We 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 is asked to answer the queries, the sum of the endurance of a consecutive part of the battleship line.

Notice that square root operation should is rounded down to integer.

Inputthe input contains several test cases, terminated by EOF.

For each test case, the first line contains a single integer n, denoting there is n battleships of evil in a line. (1 <= N <= 100000)

The second line contains N integers Ei, indicating the endurance value of all battleship from the beginning of the line T o the end. You can assume, the sum of all endurance value are 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 would decrease the endurance value of the battleships between the X -th and y-th battleship, inclusive. The t=1 denoting the commander which ask for the sum of the endurance value of the battleship between a D y-th, inclusive.

Outputfor each test case, print the case number at the first line. Then print one to 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

Sourcethe 36th ACM/ICPC Asia Regional Shanghai Site--online Contest

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#include <stdio.h> #include <string.h> #include <math.h>__int64 node[400100];void build_tr (int l,int R,int tr) {if (l==r) {scanf ("%i64d", &node[tr]); return;} Node[tr].sum=0;int m= (l+r) >>1;build_tr (l,m,tr<<1); build_tr (m+1,r,tr<<1|1); node[tr]=node[tr <<1]+NODE[TR<<1|1];} void update (int l,int r,int l,int r,int tr) {if (node[tr]==r-l+1) return;if (l==r) {node[tr]=sqrt ((double) node[tr]); return;} int m= (L+R) >>1;if (l<=m) {update (l,r,l,m,tr<<1);} if (m<r) {update (l,r,m+1,r,tr<<1|1);} NODE[TR]=NODE[TR<<1]+NODE[TR<<1|1];} __int64 query (int l,int r,int l,int r,int tr) {if (l<=l&&r<=r) {return NODE[TR]; } int m= (L+R) >>1; __int64 ans=0; if (l<=m) ans+=query (l,r,l,m,tr<<1); if (m<r) ans+=query (l,r,m+1,r,tr<<1|1); return ans; } int main () {int n,c=0;; while (scanf ("%d", &n)!=eof) {printf ("Case #%d:\n", ++c); Build_tr (1,n,1); int Q;SCANF ("%d", &q), while (q--) {int c,a,b;scanf ("%d%d%d", &c,&a,&b), if (a>b) {a^=b;b^=a;a^=b;} if (!c) {update (a,b,1,n,1);} else{printf ("%i64d\n", Query (a,b,1,n,1));}} printf ("\ n");}}

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Hdoj topic 4027 Can You answer these queries? (segment tree, interval minus square root, interval sum)