Nice boat

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 2155    Accepted Submission(s): 975 Problem Description There is an old country and the king fell in love with a devil. The devil always asks the king to do some crazy things. Although the king used to be wise and beloved by his people. Now he is just like a boy in love and can’t refuse any request from the devil. Also, this devil is looking like a very cute Loli. Let us continue our story, z*p(actually you) defeat the 'MengMengDa' party's leader, and the 'MengMengDa' party dissolved. z*p becomes the most famous guy among the princess's knight party.  One day, the people in the party find that z*p has died. As what he has done in the past, people just say 'Oh, what a nice boat' and don't care about why he died. Since then, many people died but no one knows why and everyone is fine about that. Meanwhile, the devil sends her knight to challenge you with Algorithm contest. There is a hard data structure problem in the contest: There are n numbers a_1,a_2,...,a_n on a line, everytime you can change every number in a segment [l,r] into a number x(type 1), or change every number a_i in a segment [l,r] which is bigger than x to gcd(a_i,x) (type 2). You should output the final sequence.   Input The first line contains an integer T, denoting the number of the test cases. For each test case, the first line contains a integers n. The next line contains n integers a_1,a_2,...,a_n separated by a single space. The next line contains an integer Q, denoting the number of the operations. The next Q line contains 4 integers t,l,r,x. t denotes the operation type. T<=2,n,Q<=100000 a_i,x >=0 a_i,x is in the range of int32(C++)   Output For each test case, output a line with n integers separated by a single space representing the final sequence. Please output a single more space after end of the sequence   Sample Input 1 10 16807 282475249 1622650073 984943658 1144108930 470211272 101027544 1457850878 1458777923 2007237709 10 1 3 6 74243042 2 4 8 16531729 1 3 4 1474833169 2 1 8 1131570933 2 7 9 1505795335 2 3 7 101929267 1 4 10 1624379149 2 2 8 2110010672 2 6 7 156091745 1 2 5 937186357   Sample Output 16807 937186357 937186357 937186357 937186357 1 1 1624379149 1624379149 1624379149

题意：一段区间有两个操作，把一段区间变成x，和把一段区间比x大的数和x求个gcd

lazy数组标记当前区间的子区间lazy是否相等，如果相等，那么更新当前节点就好了。

AC代码：

#include<stdio.h> #include<string.h> #include<cstdio> #include<iostream> #include<algorithm> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 const int maxn=100000+10; int maxv[maxn<<2],lazy[maxn<<2]; int gcd(int a,int b){return b?gcd(b,a%b):a;} void PushUp(int rt){ maxv[rt]=max(maxv[rt<<1],maxv[rt<<1|1]); lazy[rt]=(lazy[rt<<1]==lazy[rt<<1|1])?lazy[rt<<1]:-1; } void Pushdown(int rt){ if(lazy[rt]>-1){ maxv[rt<<1]=maxv[rt<<1|1]=lazy[rt]; lazy[rt<<1]=lazy[rt<<1|1]=lazy[rt]; lazy[rt]=-1; } } void build(int l,int r,int rt){ int m=(l+r)>>1; lazy[rt]=-1; maxv[rt]=-1; if(l==r){ scanf("%d",&maxv[rt]); lazy[rt]=maxv[rt]; return; } build(lson); build(rson); PushUp(rt); } void Updata(int L,int R,int x,int l,int r,int rt){ int m=(l+r)>>1; if(L<=l && r<=R){ lazy[rt]=x; maxv[rt]=x; return; } Pushdown(rt); if(L<=m) Updata(L,R,x,lson); if(m<R) Updata(L,R,x,rson); PushUp(rt); } void Updata2(int L,int R,int x,int l,int r,int rt){ int m=l+r>>1; if (maxv[rt]<=x) return; if(lazy[rt]>-1 && L<=l && r<=R){ lazy[rt]=gcd(lazy[rt],x); maxv[rt]=lazy[rt]; return; } Pushdown(rt); if(L<=m) Updata2(L,R,x,lson); if(m<R) Updata2(L,R,x,rson); PushUp(rt); } void print(int l,int r,int rt){ int m=l+r>>1; if(l==r){ printf("%d ",maxv[rt]); return ; } Pushdown(rt); print(lson); print(rson); } int main(){ int t,q,l,r,x,n,j; scanf("%d",&t); while(t--){ scanf("%d",&n); build(1,n,1); scanf("%d",&q); while(q--){ scanf("%d%d%d%d",&j,&l,&r,&x); if(j==1) Updata(l,r,x,1,n,1); else Updata2(l,r,x,1,n,1); } print(1,n,1); printf("\n"); } }