树状数组，一个想法是当往p注水时，认为是其容量变小了，更新时二分枚举，注意一些优化。
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 200008;
const LL INF = (LL)1 << (LL)61;
#define MS(a, num) memset(a, num, sizeof(a))
#define PB(A) push_back(A)
#define FOR(i, n) for(int i = 0; i < n; i++)
LL C[N];
LL cap[N];
LL rem[N];
int n;//注意初始化n
inline int lowbit(int x){
return x&-x;
}
inline void add(int x, LL val){
for(int i=x;i<=n;i+=lowbit(i)){
C[i] += val;
}
}
inline LL sum(int x){//求1到x的和
LL ret = 0;
for(int i=x;i>0;i-=lowbit(i)){
ret+=C[i];
}
return ret;
}
int main(){
cin>>n;
for(int i = 1;i <= n; i++){
scanf("%I64d", &cap[i]);
rem[i] = cap[i];
add(i, cap[i]);
}
cap[n + 1] = INF;
rem[n + 1] = INF;
add(n + 1, INF);
int m;
cin>>m;
int op, x,p , k;
while(m--){
scanf("%d", &op);
if(op == 1){
scanf("%d %d", &p, &x);
int l = p, r = n +1;
int ans = l;
LL s = sum(p - 1);
int start = p;
while(l < r){
int mid = (l + r)>>1;
LL s2= sum(mid) - s;
if(s2 == 0){
start = mid + 1;
}
if(s2 < x){
l = mid + 1;
}else{
r = mid;
}
}
ans = l;
LL s3 = sum(ans -1) - s;
for(int i = start; i < ans ; i++){
if(rem[i]){
add(i, -rem[i]);
rem[i] = 0;
}
}
if(x - s3){
add(ans,-( x - s3));
rem[ans] -= x - s3;
}
}else{
scanf("%d", &k);
printf("%I64d\n", cap[k] - rem[k]);
} }
return 0;
}