题意就不说了
思路：线段树，维护区间乘积。2操作就将要除的点更新为1.
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 #include<queue> 7 #include<vector> 8 #include<set> 9 #include<string> 10 #define inf 0x3f3f3f3f 11 #define LL long long 12 #define mid(a,b) ((a+b)>>1) 13 #define ll long long 14 #define maxn 110000 15 #define IN freopen("in.txt","r",stdin); 16 using namespace std; 17 18 LL mod; 19 LL num[maxn]; 20 struct Tree 21 { 22 int left,right; 23 LL pro; /*pro为区间和，可改为最大值最小值等*/ 24 }tree[maxn<<2]; /*四倍大小*/ 25 26 27 /*递归建树*/ 28 void build(int i,int left,int right) 29 { 30 tree[i].left=left; 31 tree[i].right=right; 32 33 if(left==right){ 34 tree[i].pro=num[left]%mod; 35 return ; 36 } 37 38 int mid=mid(left,right); 39 40 build(i<<1,left,mid); 41 build(i<<1|1,mid+1,right); 42 43 tree[i].pro=(tree[i<<1].pro*tree[i<<1|1].pro)%mod; 44 } 45 46 /*单点修改，d为改变量，两函数可共存*/ 47 void update(int i,int x,LL d) 48 { 49 if(tree[i].left==tree[i].right){ 50 tree[i].pro=d; 51 return; 52 } 53 54 int mid=mid(tree[i].left,tree[i].right); 55 56 if(x<=mid) update(i<<1,x,d); 57 else update(i<<1|1,x,d); 58 59 tree[i].pro=(tree[i<<1].pro*tree[i<<1|1].pro)%mod; 60 } 61 62 /*区间结果查询*/ 63 LL query(int i,int left,int right) 64 { 65 if(tree[i].left==left&&tree[i].right==right) 66 return tree[i].pro%mod; 67 68 int mid=mid(tree[i].left,tree[i].right); 69 70 if(right<=mid) return query(i<<1,left,right); 71 else if(left>mid) return query(i<<1|1,left,right); 72 else return (query(i<<1,left,mid)*query(i<<1|1,mid+1,right))%mod; 73 } 74 75 int main(int argc, char const *argv[]) 76 { 77 //IN; 78 79 int t,ca=1;scanf("%d",&t); 80 while(t--) 81 { 82 printf("Case #%d: ",ca++); 83 int n;scanf("%d %lld",&n,&mod); 84 //fill(num,num+maxn,1); 85 int type[maxn];LL val[maxn]; 86 for(int i=1;i<=n;i++){ 87 scanf("%d %lld",&type[i],&val[i]); 88 if(type[i]==1) num[i]=val[i]; 89 if(type[i]==2) num[i]=1; 90 } 91 build(1,1,n); 92 for(int i=1;i<=n;i++){ 93 if(type[i]==1){ 94 LL ans=query(1,1,i); 95 printf("%lld ",ans%mod); 96 } 97 else if(type[i]==2){ 98 update(1,val[i],1LL); 99 LL ans=query(1,1,i); 100 printf("%lld ",ans%mod); 101 } 102 } 103 104 } 105 106 return 0; 107 }