CDOJ1597-线段树好题(2017 UESTC Training for Data Structures C)
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传送门:CDOJ1597
题目大意:
给你一个长度为n的序列,m次操作,三种操作:
1.给一段区间内的每个数乘上一个非负整数。
2.给一段区间内的每个数加上一个非负整数.
3.询问一段区间的和模上P的值。
题目思路:
首先我们很好想到用线段树来维护这个区间的和,但是这里多了个乘法,加法我们很好处理
其实想一想乘法的话我们可以另外加个乘法的懒惰数组,这样在更新时照样跟新就是,这里的问题
就是加法和乘法的更新顺序,我们可以先确定更新顺序,我们先更新乘法,在更新加法,我们知道
更新乘法是如果前面有加的,我们必须先加上在乘,而这里我们可以让加的数组加上它本身的乘以
要乘的数,这样我们就可以毫不犹豫按照我们的顺序更新
AC代码:
#include<bits/stdc++.h>using namespace std;#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define ll long longconst int maxn = 1e5+100;ll p;int n;ll Tree[maxn<<2],cadd[maxn<<2],cmul[maxn<<2];void Build(int l,int r,int rt){ cadd[rt] = 0; cmul[rt] = 1; if(l==r) { scanf("%lld",&Tree[rt]); return ; } int mid = (l+r)>>1; Build(lson); Build(rson); Tree[rt] = Tree[rt<<1]+Tree[rt<<1|1]; Tree[rt]%=p;}void pushdown(int rt,ll m){ //先让加的数组加上他要乘的倍数 cadd[rt<<1]=(cadd[rt<<1]*cmul[rt]+cadd[rt])%p; cadd[rt<<1|1]=(cadd[rt<<1|1]*cmul[rt]+cadd[rt])%p; //先更新乘法 cmul[rt<<1] *= cmul[rt]; cmul[rt<<1|1] *= cmul[rt]; cmul[rt<<1] %= p; cmul[rt<<1|1] %= p; //后更新加法 Tree[rt<<1]=(Tree[rt<<1]*cmul[rt]+(m-m/2ll)*cadd[rt])%p; Tree[rt<<1|1]=(Tree[rt<<1|1]*cmul[rt]+(m/2ll)*cadd[rt])%p; cadd[rt] = 0; cmul[rt] = 1;}void updata(int L,int R,int l,int r,int rt ,ll c,int op){ if(L<=l&&r<=R) { if(op==1) { cadd[rt] = cadd[rt]*c; cadd[rt]%=p; cmul[rt]*=c; cmul[rt]%=p; Tree[rt]*=c; Tree[rt]%=p; } else { cadd[rt]+=c; cadd[rt]%=p; Tree[rt]+=c*(r-l+1ll); Tree[rt]%=p; } return ; } pushdown(rt,r-l+1ll); int mid = (l+r)>>1; if(L<=mid) { updata(L,R,lson,c,op); } if(R>mid) { updata(L,R,rson,c,op); } Tree[rt] = (Tree[rt<<1]+Tree[rt<<1|1])%p;}ll quary(int L,int R,int l,int r,int rt){ if(L<=l&&r<=R)return Tree[rt]%p; pushdown(rt,r-l+1ll); int mid = (l+r)>>1; ll res = 0; if(L<=mid)res+=quary(L,R,lson); if(R>mid)res+=quary(L,R,rson); return res%p;}int main(){ cin>>n>>p; Build(1,n,1); int m; cin>>m; while(m--) { int id,a,b; ll c; scanf("%d",&id); if(id==1) { scanf("%d%d%lld",&a,&b,&c); updata(a,b,1,n,1,c,1); } else if(id==2) { scanf("%d%d%lld",&a,&b,&c); updata(a,b,1,n,1,c,2); } else { scanf("%d%d",&a,&b); printf("%lld\n",quary(a,b,1,n,1)); } } return 0;}
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