Light oj 1112 - Curious Robin Hood【单点更新】

1112 - Curious Robin Hood

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Time Limit: 1 second(s)Memory Limit: 64 MB

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

Output for 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

Case 1:

5

14

1

13

2

Notes

Dataset is huge, use faster I/O methods.

题意：

给出一组数，进行题目给出的三种操作，输出需要的结果。

题解：

最基础的线段树题目，只需要单点更新和区间查询

很久没做线段树了，没想到还能 1 A....这是题目有多水....

/*http://blog.csdn.net/liuke19950717*/#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const int maxn=1e5+5;int x[maxn],sumtree[maxn*4],kase=0;void build(int rt,int l,int r){if(l==r){sumtree[rt]=x[l];return;}int mid=(l+r)>>1,tp=rt<<1;build(tp,l,mid);build(tp|1,mid+1,r);sumtree[rt]=sumtree[tp]+sumtree[tp|1];}void update(int rt,int l,int r,int a,int b){if(l==r){sumtree[rt]=x[l]=b;return;}int mid=(l+r)>>1,tp=rt<<1;if(mid>=a){update(tp,l,mid,a,b);}else{update(tp|1,mid+1,r,a,b);}sumtree[rt]=sumtree[tp]+sumtree[tp|1];}int find(int rt,int l,int r,int a,int b){if(l>=a&&r<=b){return sumtree[rt];}int mid=(l+r)>>1,tp=rt<<1,ans=0;if(mid>=a){ans+=find(tp,l,mid,a,b);}if(mid<b){ans+=find(tp|1,mid+1,r,a,b);}return ans;}int main(){//freopen("shuju.txt","r",stdin);int t;scanf("%d",&t);while(t--){int n,m;memset(sumtree,0,sizeof(sumtree));scanf("%d%d",&n,&m);for(int i=1;i<=n;++i){scanf("%d",&x[i]);}build(1,1,n);printf("Case %d:\n",++kase);while(m--){int num;scanf("%d",&num);if(num==1){int a;scanf("%d",&a);printf("%d\n",x[a+1]);update(1,1,n,a+1,0);}else if(num==2){int a,b;scanf("%d%d",&a,&b);update(1,1,n,a+1,x[a+1]+b);}else{int a,b;scanf("%d%d",&a,&b);printf("%d\n",find(1,1,n,a+1,b+1));}}}return 0;}