洛谷P3372 【模板】线段树 1 分块

#include<cstdio>
#include<cstring>
#include<cmath>

#define ll long long

using namespace std;

ll c[100005];
ll addv[320];
ll sum[320];
int n,m,g,p=1,ks=1;  //g每块大小 p临时指针 ks块数 
int f,x,y,pl,pr;  //pl左指针 pr右指针 
ll k,ans;

int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)scanf("%lld",&c[i]);
    memset(addv,0,sizeof(addv));
    g=sqrt(n);
    for(int i=1;i<=n;i++){
        if(p>g){
            p=1;
            ks++;
        }
        sum[ks]+=c[i];
        p++;
    }
    for(int i=1;i<=m;i++){
        scanf("%d",&f);
        if(f==1){
            scanf("%d%d%lld",&x,&y,&k);
            if(x%g)pl=(x/g)+1;else pl=x/g;
            if(y%g)pr=(y/g)+1;else pr=y/g;
            if(x!=(pl-1)*g+1){
                while(x<=pl*g){
                    c[x]+=k;
                    sum[pl]+=k;
                    x++;
                }
                pl++;
            }
            if(y!=pr*g){
                while(y>(pr-1)*g){
                    c[y]+=k;
                    sum[pr]+=k;
                    y--;
                }
                pr--;
            }
            for(int j=pl;j<=pr;j++){
                addv[j]+=k;
                sum[j]+=k*g;
            }
        }
        else{
            scanf("%d%d",&x,&y);
            if(x%g)pl=(x/g)+1;else pl=x/g;
            if(y%g)pr=(y/g)+1;else pr=y/g;
            ans=0;
            if(x!=(pl-1)*g+1){
                while(x<=pl*g){
                    ans+=c[x]+addv[pl];
                    x++;
                }
                pl++;
            }
            if(y!=pr*g){
                while(y>(pr-1)*g){
                    ans+=c[y]+addv[pr];
                    y--;
                }
                pr--;
            }
            for(int j=pl;j<=pr;j++)ans+=sum[j];
            printf("%lld
",ans);
        }
    }
    
    return 0;
}