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考点:树状数组的应用,如果懂树状数组,那么这个题很轻松拿下
#include<iostream>
#include<sstream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<ctime>
#include<algorithm>
using namespace std;
#define _for(i,lo,hi) for(int i=(lo);i<(hi);i++)
#define lowbit(i) (i&(-i))
int n;
int A[100005];//开始时把这里也定义为long long类型,所以最后一个算例过不了,只能得90
long long C[100005];
//A[x]加上v
void updata(int x,int v){
if(v==0)return;
for(int i=x;i<n+1;i+=lowbit(i)){
C[i]+=v;
}
}
//计算A[1]-A[x]的和
long long getsum(int x){
long long sum=0;
for(int i=x;i>0;i-=lowbit(i)){
sum+=C[i];
}
return sum;
}
int main() {
//freopen("D:\\input.txt","r",stdin);
int m,opt;
scanf("%d %d",&n,&m);
memset(C,0,sizeof(C));
//预处理,生成树状数组
_for(i,1,n+1){scanf("%d",&A[i]);updata(i,A[i]);}
//在线查询
int l,r,v;
_for(i,0,m){
scanf("%d",&opt);
if(opt==1){
scanf("%d %d %d",&l,&r,&v);//本来50分
if(v==1)continue;//加这一行优化,80分
_for(j,l,r+1){
if(A[j]>=v&&A[j]%v==0){//加上A[i]>=v90分
updata(j,A[j]/v-A[j]);
A[j]/=v;
}
}
}else{
scanf("%d %d",&l,&r);
printf("%lld\n",getsum(r)-getsum(l-1));
}
}
return 0;
}