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 integer n, means the number of level. (1 <= n <= 100000). The second line follows n integers, the ith integer means the number in level i. The third line is an integer k, 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 3
3
q 2
p 1 2
q 2
Sample Output
21
Author: Lin, Yue
Source: ZOJ Monthly, December 2009
/* ZOJ 3279 */ #include <iostream> #include <cstdio> #include <cstring> using namespace std; const int maxn=100000; int Tree[100010]; int k[100010]; int lowbit(int x) { return -x&x; } void add(int x,int val) { for(int i=x;i<=maxn;i+=lowbit(i)) Tree } int get(int x) { int sum=0; for(int i=x;i;i-=lowbit(i)) sum+=Tree return sum; } int n,m; int main() { while(scanf("%d",&n)!=EOF) { memset(Tree,0,sizeof(Tree)); for(int i=1;i<=n;i++) { scanf("%d",&k add(i,k } scanf("%d",&m); char ch[2];int a,b; while(m--) { scanf("%s",ch); if(ch[0]=='p') { scanf("%d%d",&a,&b); add(a,-k[a]); add(a,b); k[a]=b; } if(ch[0]=='q') { scanf("%d",&a); int high=n,low=1; int ans=n; while(low<=high) { int mid=(high+low)/2; if(get(mid)>=a) { high=mid-1; ans=mid; } else { low=mid+1; } } printf("%d ",ans); } } } return 0; } |