ZOJ 3279 Ants（线段树）

Ants

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Time Limit: 2 Seconds      Memory Limit: 32768 KB

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echo is a curious and clever girl, and she is addicted to the ants recently.

She knows that the ants are divided into many levels depends on ability, also, she finds the number of each level will change.

Now, she will give two kinds of operations as follow :

First, "p a b", the number of ants in level a change to b.

Second, "q x", it means if the ant's ability is rank xth in all ants, what level will it in?

Input

There are multi-cases, and you should use EOF to check whether it is in the end of the input. The first line is an integern, means the number of level. (1 <= n <= 100000). The second line followsn integers, the ith integer means the number in level i. The third line is an integerk, means the total number of operations. Then following k lines, each line will be"p a b" or "q x", and 1 <= x <= total ants, 1 <= a <=n, 0 <= b. What's more, the total number of ants won't exceed 2000000000 in any time.

Output

Output each query in order, one query each line.

Sample Input

31 2 33q 2p 1 2q 2

Sample Output

21

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Author: Lin, Yue

Source: ZOJ Monthly, December 2009

给出每个等级的数目，后面m个操作，q查询排名x在第几个等级，p a b,把a等级的数目改为b

ac代码

#include<stdio.h>#include<string.h>int node[100010<<2];void build(int l,int r,int tr){if(l==r){scanf("%d",&node[tr]);return;}int mid=(l+r)>>1;build(l,mid,tr<<1);build(mid+1,r,tr<<1|1);node[tr]=node[tr<<1]+node[tr<<1|1];}void update(int pos,int val,int l,int r,int tr){if(l==r){node[tr]=val;return;}int mid=(l+r)>>1;if(pos<=mid)update(pos,val,l,mid,tr<<1);elseupdate(pos,val,mid+1,r,tr<<1|1);node[tr]=node[tr<<1]+node[tr<<1|1];}int query(int val,int l,int r,int tr){if(l==r){return l;}int mid=(l+r)>>1;if(val<=node[tr<<1])query(val,l,mid,tr<<1);elsequery(val-node[tr<<1],mid+1,r,tr<<1|1);}int main(){int n;while(scanf("%d",&n)!=EOF){int i;memset(node,0,sizeof(node));build(1,n,1);int m;scanf("%d",&m);while(m--){char s[2];scanf("%s",s);if(s[0]=='q'){int x;scanf("%d",&x);int ans=query(x,1,n,1);printf("%d\n",ans);}else{int a,b;scanf("%d%d",&a,&b);update(a,b,1,n,1);}}}}