远古div1D题，线段树维护区间取模，单点修改和区间和
有一个神秘结论：x%p<x/2(p<x)
这意味着取模次数较多的时候x会以指数形式减小，于是考虑暴力区间取模，顺便维护一个区间最大值，如果区间最大值小于模数直接return即可
注意一个问题，不能直接把区间和取模，每个点取模和区间和取模是两回事
代码：
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} int n,m,op,l,r,k,x; int a[maxn],mx[maxn*20],sum[maxn*20]; void up(int p){ mx[p]=max(mx[ll(p)],mx[rr(p)]); sum[p]=sum[ll(p)]+sum[rr(p)]; } void build(int p,int l,int r){ if(l==r){ sum[p]=mx[p]=a[l]; return; } int m=mm(l,r); build(ll(p),l,m); build(rr(p),m+1,r); up(p); } void setx(int p,int l,int r,int k,int x){ if(l==r&&l==k){ mx[p]=sum[p]=x; return; } int m=mm(l,r); if(k<=m) setx(ll(p),l,m,k,x); else setx(rr(p),m+1,r,k,x); up(p); } void modx(int p,int l,int r,int L,int R,int x){ if(mx[p]<x) return; if(l==r){ sum[p]%=x; mx[p]%=x; return; } int m=mm(l,r); if(L<=m) modx(ll(p),l,m,L,R,x); if(R>m) modx(rr(p),m+1,r,L,R,x); up(p); } int ask(int p,int l,int r,int L,int R){ if(L<=l&&r<=R) return sum[p]; int m=mm(l,r),res=0; if(L<=m) res+=ask(ll(p),l,m,L,R); if(R>m) res+=ask(rr(p),m+1,r,L,R); return res; } signed main(){ scc(n,m); re(i,1,n) sc(a[i]); build(1,1,n); re(i,1,m){ sc(op); if(op==1) scc(l,r),pr(ask(1,1,n,l,r)),prn(); else if(op==2) sccc(l,r,x),modx(1,1,n,l,r,x); else if(op==3) scc(k,x),setx(1,1,n,k,x); } return 0; } /* 5 100 1 2 3 4 5 2 3 5 4 3 3 5 1 2 5 2 1 3 3 1 1 3 */