08: Challenge 1,08 challenge

08: Challenge 1,08 challenge

Total time limit:

10000 ms

 

Time limit for a single test point:

1000 ms

 

Memory limit:

262144kB

Description

For an N-length series with M operations, each operation is one of the following two types:

(1) modify a number in a series

(2) evaluate the value of a position in a sequence after an operation

Input

The first line has two integers, N and M.
The N integers in the second row represent the series.
Next, the M line starts with one character. If the character is 'M', it indicates a modification operation. The next two integers x and y, it indicates that the value at the position x is changed to y. If the character is 'Q', it indicates a query operation. The next two integers x and y are involved, returns the value of x after the y operation.

Output

Output a single line for each query operation, indicating the answer.

Sample Input

5 31 2 3 4 5Q 1 0M 1 3Q 1 2

Sample output

13

Prompt

1 <= N <= 10 ^ 5, 1 <= M <= 10 ^ 5, the input is valid, and All integers can be stored with 32-bit integers.

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• Question

At first glance, many people thought it was necessary to use the Chairman tree or something.

However.

Rope !!.

There is nothing to explain, that is, the template question cannot be left empty ,,,

 

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<ext/rope>using namespace std;using namespace __gnu_cxx;const int MAXN=2000050;const int maxn=0x7fffffff;void read(int &n){    char c='+';int x=0;bool flag=0;    while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}    while(c>='0'&&c<='9'){x=x*10+(c-48);c=getchar();}    flag==1?n=-x:n=x;}rope<int> *rp[MAXN];int a[MAXN];int tot=0;int main(){ios::sync_with_stdio(0);    int n,m;    read(n);read(m);    for(int i=1;i<=n;i++)    read(a[i]);    rp[0]=new rope<int>(a+1,a+n+1);    for(int i=1;i<=m;i++)    {    rp[i]=new rope<int>(*rp[i-1]);    char c=getchar();int x,y;    if(c=='Q')    {    int l,r;    int ans=0;    read(l);read(r);read(x);    for(int i=l;i<=r;i++)    ans+=(rp[x]->at(i-1));    printf("%d\n",ans);}else{read(x);read(y);rp[i]->replace(x-1,y);}}    return 0;}