CF438D The Child and Sequence(线段树区间取模)
CF438D The Child and Sequence(线段树区间取模)
思路
维护区间的最大值，取模的时候如果最大值小于模数就跳过此区间；否则，对每个点暴力修改。由于每次取模后 xmodp<2/x，所以每个点最多被修改logx次。
代码
// Problem: D. The Child and Sequence// Contest: Codeforces - Codeforces Round #250 (Div. 1)// URL: https://codeforces.com/problemset/problem/438/D// Memory Limit: 256 MB// Time Limit: 4000 ms// // Powered by CP Editor (https://cpeditor.org)#include<bits/stdc++.h>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair<ll, ll>PLL;typedef pair<int, int>PII;typedef pair<double, double>PDD;#define I_int llinline ll read(){
ll x = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9')
{ if(ch == '-')f = -1;
ch = getchar();
} while(ch >= '0' && ch <= '9')
{
x = x * 10 + ch - '0';
ch = getchar();
} return x * f;
}
inline void out(ll x){ if (x < 0) x = ~x + 1, putchar('-'); if (x > 9) out(x / 10); putchar(x % 10 + '0');
}
inline void write(ll x){ if (x < 0) x = ~x + 1, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0');
}
#define read read()#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)#define rep(i,a,b) for(int i=(a);i<=(b);i++)#define repp(i,a,b) for(int i=(a);i<(b);i++)#define per(i,a,b) for(int i=(a);i>=(b);i--)#define perr(i,a,b) for(int i=(a);i>(b);i--)ll ksm(ll a, ll b){
ll res = 1; while(b)
{ if(b & 1)res = res * a ;
a = a * a ;
b >>= 1;
} return res;
}const int inf = 0x3f3f3f3f;const int maxn=1e5+7,maxm=210000;int n,m,a[maxn];struct node{
int l,r;
ll sum,maxx;
}tr[maxn<<2];void pushup(int u){
tr[u].sum=tr[u<<1].sum+tr[u<<1|1].sum;
tr[u].maxx=max(tr[u<<1].maxx,tr[u<<1|1].maxx);
}void build(int u,int l,int r){
tr[u]={l,r}; if(l==r){
tr[u].sum=tr[u].maxx=a[l]; return ;
} int mid=(l+r)/2;
build(u<<1,l,mid);build(u<<1|1,mid+1,r);
pushup(u);
}ll query_sum(int u,int l,int r){ if(tr[u].l>=l&&tr[u].r<=r) return tr[u].sum; int mid=(tr[u].l+tr[u].r)/2;
ll res=0; if(l<=mid) res+=query_sum(u<<1,l,r); if(r>mid) res+=query_sum(u<<1|1,l,r); return res;
}void update_mod(int u,int l,int r,ll x){ if(tr[u].l>=l&&tr[u].r<=r&&tr[u].maxx<x) return ; if(tr[u].l>=l&&tr[u].r<=r&&tr[u].l==tr[u].r){
tr[u].maxx%=x;tr[u].sum%=x; return ;
} int mid=(tr[u].l+tr[u].r)/2; if(l<=mid) update_mod(u<<1,l,r,x); if(r>mid) update_mod(u<<1|1,l,r,x);
pushup(u);
}void update_pos(int u,int l,int r,int x){ if(tr[u].l==l&&tr[u].r==r){
tr[u].sum=tr[u].maxx=x; return ;
} int mid=(tr[u].l+tr[u].r)/2; if(l<=mid) update_pos(u<<1,l,r,x); if(r>mid) update_pos(u<<1|1,l,r,x);
pushup(u);
}int main(){
n=read,m=read;
rep(i,1,n) a[i]=read;
build(1,1,n); while(m--){ int op=read; if(op==1){ int l=read,r=read; printf("%lld\n",query_sum(1,l,r));
} else if(op==2){ int l=read,r=read;
ll x=read;
update_mod(1,l,r,x);
} else{ int k=read,x=read;
update_pos(1,k,k,x);
}
}
return 0;
}