（線段樹 點更新 區間求和）lightoj1112

鏈接：

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88230#problem/D （密碼0817）

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Description

Robin Hood likes to loot rich people since he helps the poor people with this money. Instead of keeping all the money together he does another trick. He keeps?n?sacks where he keeps this money. The sacks are numbered from?0?to?n-1.

Now each time?he can he can do one of the three tasks.

1)??????????????????Give all the money of the?i^th?sack to the poor, leaving the sack empty.

2)??????????????????Add new amount (given in input) in the?i^th?sack.

3)??????????????????Find the total amount of money from?i^th?sack to?j^th?sack.

Since he is not a?programmer, he seeks your help.

Input

Input starts with an integer?T (≤ 5), denoting the number of test cases.

Each case contains two integers?n (1 ≤ n ≤ 10⁵)?and?q (1 ≤ q ≤ 50000). The next line contains?n?space separated integers in the range?[0, 1000]. The?i^th?integer denotes the initial amount of money in the?i^th?sack?(0 ≤ i < n).

Each of the next?q?lines contains a task in one of the following form:

1 i????????Give all the money of the?i^th(0 ≤ i < n)?sack to the poor.

2 i v?????Add money?v (1 ≤ v ≤ 1000)?to the?i^th(0 ≤ i < n)?sack.

3 i j??????Find the total amount of money from?i^th?sack to?j^th?sack?(0?≤ i ≤ j < n).

Output

For each test case, print the case number first. If the query type is?1, then print the amount of money given to the poor. If the query type is?3, print the total amount from?i^th?to?j^th?sack.

Sample Input

1

5 6

3 2 1 4 5

1 4

2 3 4

3 0 3

1 2

3 0 4

1 1

Sample Output

Case 1:

5

14

1

13

2

?

?

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代碼：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;#define Lson (r<<1)
#define Rson (r<<1|1)
#define Mid e[r].mid()const int N = 100005;struct node
{int L, R, sum;int mid(){return (L+R)/2;}
} e[N<<2];int a[N], sum;void BuildTree(int r, int L, int R)
{e[r].L = L , e[r].R = R;if(L==R){e[r].sum = a[L];return ;}BuildTree(Lson, L, Mid);BuildTree(Rson, Mid+1, R);e[r].sum = e[Lson].sum + e[Rson].sum;
}void Oper(int r, int i, int w)
{if(e[r].L == e[r].R){e[r].sum = w;return ;}if(i<=Mid)Oper(Lson, i, w);elseOper(Rson, i, w);e[r].sum = e[Lson].sum + e[Rson].sum;
}int Query(int r, int L, int R)
{if(e[r].L==L && e[r].R==R)return e[r].sum;if(R<=Mid)return Query(Lson, L, R);else if(L>Mid)return Query(Rson, L, R);else{int LL = Query(Lson, L, Mid);int RR = Query(Rson, Mid+1, R);return LL + RR;}}int main()
{int t, n, m, iCase=1;scanf("%d", &t);while(t--){int i, L, R, w, x;scanf("%d%d", &n, &m);for(i=1; i<=n; i++)scanf("%d", &a[i]);BuildTree(1, 1, n);printf("Case %d:\n", iCase++);while(m--){scanf("%d", &x);if(x==3){scanf("%d%d", &L, &R);sum = 0;L++, R++;printf("%d\n", Query(1, L, R));}else{scanf("%d", &i);i++;if(x==1){printf("%d\n", a[i]);a[i] = 0;}else{scanf("%d", &w);a[i] += w;}Oper(1, i, a[i]); ///由于都是點的操作，可以直接操做后在去操作樹，感覺和直接操作樹是一樣的，不過對于///這題來說，這種似乎更好些， 因為有加和清零的，對點的操作是不一樣的， 然而區間查詢就很簡單了，不說了}}}return 0;
}

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