题目链接：http://www.bnuoj.com/bnuoj/problem_show.php?pid=29359
题意：一个数列，有三种操作：
1.区间[a,b]之间大于零的数整出c。
2.区间[a,b]之间所有的数减去c。
3.求区间[a,b]的和。
只要注意到每个数最多除lgn次，总共除n*lgn次，那么直接对除法进行单点更新就可了，关键要分析好复杂度。。
//STATUS:C++_AC_3020MS_33996KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef long long LL;
typedef unsigned long long ULL;
//const
const int N=;
const int INF=0x3f3f3f3f;
const int MOD=,STA=;
const LL LNF=1LL<<;
const double EPS=1e-;
const double OO=1e15;
const int dx[]={-,,,};
const int dy[]={,,,-};
const int day[]={,,,,,,,,,,,,};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End LL sum[N<<],miu[N<<];
int flag[N<<],num[N];
int T,n,m,a,b,c;
LL ans; void pushdown(int l,int r,int rt)
{
if(miu[rt]){
miu[rt<<]+=miu[rt];
miu[rt<<|]+=miu[rt];
sum[rt]+=miu[rt]*(r-l+);
miu[rt]=;
}
} void pushup(int l,int r,int rt)
{
int mid=(l+r)>>;
sum[rt]=sum[rt<<]+(mid-l+)*miu[rt<<]
+sum[rt<<|]+(r-mid)*miu[rt<<|];
flag[rt]=flag[rt<<]|flag[rt<<|];
} void build(int l,int r,int rt)
{
miu[rt]=;
if(l==r){
sum[rt]=num[l];
flag[rt]=num[l]>;
return;
}
int mid=(l+r)>>;
build(lson);
build(rson);
sum[rt]=sum[rt<<]+sum[rt<<|];
flag[rt]=flag[rt<<]|flag[rt<<|];
} void update1(int l,int r,int rt)
{
if(l==r){
sum[rt]+=miu[rt];
miu[rt]=;
sum[rt]=(sum[rt]>?sum[rt]/=c:sum[rt]);
flag[rt]=sum[rt]>;
return;
}
int mid=(l+r)>>;
pushdown(l,r,rt);
if(a<=mid && flag[rt<<])update1(lson);
if(b>mid && flag[rt<<|])update1(rson);
pushup(l,r,rt);
} void update2(int l,int r,int rt)
{
if(a<=l && r<=b){
miu[rt]-=c;
return;
}
int mid=(l+r)>>;
pushdown(l,r,rt);
if(a<=mid)update2(lson);
if(b>mid)update2(rson);
pushup(l,r,rt);
} void query(int l,int r,int rt)
{
if(a<=l && r<=b){
ans+=sum[rt]+(r-l+)*miu[rt];
return;
}
int mid=(l+r)>>;
pushdown(l,r,rt);
if(a<=mid)query(lson);
if(b>mid)query(rson);
pushup(l,r,rt);
} int main()
{
// freopen("in.txt","r",stdin);
int i,j,ca=;
char s[];
scanf("%d",&T);
while(T--)
{
printf("Case %d:\n",ca++);
scanf("%d%d",&n,&m);
mem(sum,),mem(miu,);
for(i=;i<=n;i++){
scanf("%d",&num[i]);
}
build(,n,);
while(m--){
scanf("%s",s);
if(s[]=='D'){
scanf("%d%d%d",&a,&b,&c);
if(c==)continue;
update1(,n,);
}
else if(s[]=='M'){
scanf("%d%d%d",&a,&b,&c);
update2(,n,);
}
else {
scanf("%d%d",&a,&b);
ans=;
query(,n,);
printf("%lld\n",ans);
}
}
putchar('\n');
}
return ;
}