LOJ #6280. Introduction to series partitioning 4,

LOJ #6280. Introduction to series partitioning 4,
Memory limit: 256 MiB time limit: 500 ms standard input/output question type: traditional evaluation method: text comparison uploaded by: hzwer submit Record Statistics Discussion test data question description

A series with a length of nnn and nnn operations are provided. The operations involve interval addition and interval summation.

Input Format

Enter a number nnn in the first line.

Enter nnn numbers in the second row. the I-th digit is aia_iai, separated by spaces.

Next, enter the nnn line and enter four numbers in each line: opt \ mathrm {opt} opt, lll, rrr, and ccc, separated by spaces.

If opt = 0 \ mathrm {opt} = 0opt = 0, the numbers between [l, r] [l, r] [l, r] are added with ccc.

If opt = 1 \ mathrm {opt} = 1opt = 1, it indicates that the query is located in [l, r] [l, r] [l, r] And mod (c + 1) mod (c + 1) mod (c + 1) mod (c + 1) of all numbers ).

Output Format

For each query, a row of numbers is output to indicate the answer.

Sample Input

41 2 2 30 1 3 11 1 4 40 1 2 21 1 2 4

Sample output

14

Data range and prompt

For data of 100% 100 \ % 100%, 1 ≤ n ≤ 50000, − 231 ≤ others 1 \ leq n \ leq 50000, -2 ^ {31} \ leq \ mathrm {others} 1 ≤ n ≤ 50000, − 231 ≤ others, ans ≤ 231 − 1 \ mathrm {ans} \ leq 2 ^ {31}-1ans ≤ 231 − 1.

 

Multipart

Maintain the Block Value

The rest follow the routine

 

 

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<vector>#define int long long using namespace std;const int MAXN=2*1e6+10;const int INF=1e8+10;inline char nc(){    static char buf[MAXN],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++;}inline int read(){    char c=nc();int x=0,f=1;    while(c<'0'||c>'9'){if(c=='-')f=-1;c=nc();}    while(c>='0'&&c<='9'){x=x*10+c-'0';c=nc();}    return x*f;}int a[MAXN],belong[MAXN],L[MAXN],R[MAXN],sum[MAXN],tag[MAXN],block;void IntervalAdd(int l,int r,int val){    for(int i=l;i<=min(R[l],r);i++) a[i]+=val,sum[belong[i]]+=val;    if(belong[l]!=belong[r])        for(int i=L[r];i<=r;i++)            a[i]+=val,sum[belong[i]]+=val;    for(int i=belong[l]+1;i<=belong[r]-1;i++)        tag[i]+=val;}int Query(int l,int r,int mod){    int ans=0;    for(int i=l;i<=min(R[l],r);i++)         ans+=a[i]+tag[belong[i]],ans=ans%mod;        if(belong[l]!=belong[r])        for(int i=L[r];i<=r;i++)            ans+=a[i]+tag[belong[i]];    for(int i=belong[l]+1;i<=belong[r]-1;i++)        ans+=sum[i]+tag[i]*block,ans=ans%mod;    return ans%mod;}main(){    #ifdef WIN32    freopen("a.in","r",stdin);   // freopen("b.out","w",stdout);    #else    #endif    int N=read();block=sqrt(N);    for(int i=1;i<=N;i++) belong[i]=(i-1)/block+1,L[i]=(belong[i]-1)*block+1,R[i]=belong[i]*block;    for(int i=1;i<=N;i++) a[i]=read();    for(int i=1;i<=N;i++) sum[belong[i]]+=a[i];    for(int i=1;i<=N;i++)    {        int opt=read(),l=read(),r=read(),c=read();        if(opt==0) IntervalAdd(l,r,c);        else                    printf("%lld\n",Query(l,r,c+1));     }     return 0;}