B. Sereja and Array
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Sereja has got an array, consisting of n integers, a1, a2, ..., an. Sereja is an active boy, so he is now going to complete m operations. Each operation will have one of the three forms:
Make vi-th array element equal to xi. In other words, perform the assignment avi = xi.
Increase each array element by yi. In other words, perform n assignments ai = ai + yi (1 ≤ i ≤ n).
Take a piece of paper and write out the qi-th array element. That is, the element aqi.
Help Sereja, complete all his operations.
Input
The first line contains integers n, m (1 ≤ n, m ≤ 105). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the original array.
Next m lines describe operations, the i-th line describes the i-th operation. The first number in the i-th line is integer ti (1 ≤ ti ≤ 3) that represents the operation type. If ti = 1, then it is followed by two integers vi and xi, (1 ≤ vi ≤ n, 1 ≤ xi ≤ 109). If ti = 2, then it is followed by integer yi (1 ≤ yi ≤ 104). And if ti = 3, then it is followed by integer qi (1 ≤ qi ≤ n).
Output
For each third type operation print value aqi. Print the values in the order, in which the corresponding queries follow in the input.
Sample test(s)
input
10 11
1 2 3 4 5 6 7 8 9 10
3 2
3 9
2 10
3 1
3 10
1 1 10
2 10
2 10
3 1
3 10
3 9
output
2
9
11
20
30
40
39
水題，統計與峰頂的距離，模擬
[cpp]
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define MAXM (1000000+10)
int n,m;
/*
struct tree_arr
{
int a[MAXM*4];
tree_arr(){memset(a,0,sizeof(a));}
void inc(int x,int c)
{
for(int i=x;i<=n;i+=i&(-i)) a[x]+=c;
}
int qur(int x)
{
int ans=0;
for(int i=x;i;i-=i&(-i)) ans+=a[x];
return ans;
}
}T1,T2;
void Ins()
{
}*/
int a[MAXM],tmp=0;
int main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
cin>>n>>m;
For(i,n) {scanf("%d",&a[i]);}
For(i,m)
{
int type;
scanf("%d",&type);
if (type==1)
{
int v,x;
scanf("%d%d",&v,&x);
a[v]+=x-(a[v]+tmp);
}
else if (type==2)
{
int w;
cin>>w;
tmp+=w;
}
else
{
int v;
scanf("%d",&v);
cout<<a[v]+tmp<<endl;
}
}
return 0;
}
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define MAXM (1000000+10)
int n,m;
/*
struct tree_arr
{
int a[MAXM*4];
tree_arr(){memset(a,0,sizeof(a));}
void inc(int x,int c)
{
for(int i=x;i<=n;i+=i&(-i)) a[x]+=c;
}
int qur(int x)
{
int ans=0;
for(int i=x;i;i-=i&(-i)) ans+=a[x];
return ans;
}
}T1,T2;
void Ins()
{
}*/
int a[MAXM],tmp=0;
int main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
cin>>n>>m;
For(i,n) {scanf("%d",&a[i]);}
For(i,m)
{
int type;
scanf("%d",&type);
if (type==1)
{
int v,x;
scanf("%d%d",&v,&x);
a[v]+=x-(a[v]+tmp);
}
else if (type==2)
{
int w;
cin>>w;
tmp+=w;
}
else
{
int v;
scanf("%d",&v);
cout<<a[v]+tmp<<endl;
}
}
return 0;
}
