给一棵树，树上每个节点有一个权值，有两个操作，RAND操作查询u的子树乘积是多少以及有多少因数，SEED操作把节点u乘上v
n，q <= 1e5。数值小于等于1e9，最大的质因数不超过13
组队训练和队友一起写的，写到头昏，代码也是合力完成的，我数学几乎为0，数学部分感谢队友@lllrj抬一手
思路：首先由于乘积的值过大，线段树不能直接维护权值，考虑到查询的是因数个数，所以线段树直接维护2,3,5,7,11,13的指数，乘一个数就是把这个数质因数分解再单点更新，查询就是区间求和，乘积用快速幂计算，至于因数的个数由唯一分解定理可知，把一个数分解质因数变成(x1^n1)*(x2^n2)*(x3^n3)......的形式，因数的个数就是(n1+1)*(n2+1)*(n3+1).....然后两个结果都要对1e9+7取模，比赛时后一个结果忘记取模浪费了快一个小时。。。
建树的话就是按dfs序建树，也是个常见套路，参见HDU-3974。需要注意按dfs序建树之后不能直接用编号的权值去建（比如a【3】不一定在线段树【3,3】这个点上），所以是先建树，输入的时候做更新
AC代码：217ms，29.4MB
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 #include<vector> 6 #include<queue> 7 #include<cmath> 8 #include<set> 9 #define lid id<<1 10 #define rid id<<1|1 11 #define INF 0x3f3f3f3f 12 #define LL long long 13 #define debug(x) cout << "[" << x << "]" << endl 14 using namespace std; 15 const int maxn = 1e5+5; 16 int b[]={2,3,5,7,11,13}; 17 int d[maxn][6]={0}; 18 void cal(int x, int *d) 19 { 20 for(int i = 0;i < 6 ;i++){ 21 while(x%b[i]==0)x/=b[i],d[i]++; 22 } 23 } 24 const int mx = 1e5+10; 25 const int mod = 1e9+7; 26 int L[mx], R[mx], p[mx]; 27 struct tree{ 28 int l, r; 29 int p[6]; //2,3,5,7,11,13 30 int lazy[6]; 31 }tree[mx<<2]; 32 vector<int> G[mx]; 33 int cnt; 34 LL Ans[6] = {0}; 35 36 LL qpow(LL x, LL n){ //x^n 37 LL res = 1; 38 while (n > 0){ 39 if (n & 1) res = res*x%mod; 40 x = x*x % mod; 41 n >>= 1; 42 } 43 return res; 44 } 45 46 void push_up(int id){ 47 for (int i = 0; i < 6; i++) 48 tree[id].p[i] = tree[lid].p[i]+tree[rid].p[i]; 49 } 50 51 void build(int l, int r, int id){ 52 tree[id].l = l; 53 tree[id].r = r; 54 for (int i = 0; i < 6; i++) tree[id].p[i] = 0; 55 if (l == r) return; 56 int mid = (l+r) >> 1; 57 build(l, mid, lid); 58 build(mid+1, r, rid); 59 } 60 61 void dfs(int u){ 62 L[u] = ++cnt; 63 int len = G[u].size(); 64 for (int i = 0; i < len; i++){ 65 int v = G[u][i]; 66 dfs(v); 67 } 68 R[u] = cnt; 69 } 70 71 void upd(int c, int id, int *x){ 72 if (tree[id].l == c && tree[id].r == c){ 73 for (int i = 0; i < 6; i++) 74 tree[id].p[i] += x[i]; 75 return; 76 } 77 int mid = (tree[id].l + tree[id].r)>>1; 78 if (c <= mid) upd(c, lid, x); 79 else upd(c, rid, x); 80 push_up(id); 81 } 82 83 void query(int l, int r, int id){ 84 if (tree[id].l == l && tree[id].r == r){ 85 for (int i = 0; i < 6; i++) 86 Ans[i] += tree[id].p[i]; 87 return; 88 } 89 int mid = (tree[id].l + tree[id].r)>>1; 90 if (r <= mid) query(l, r, lid); 91 else if (mid < l) query(l, r, rid); 92 else { 93 query(l, mid, lid); 94 query(mid+1, r, rid); 95 } 96 } 97 98 int main(){ 99 int n, u, v, a, q; 100 cnt = 0; 101 scanf("%d", &n); 102 for (int i = 1; i < n; i++){ 103 scanf("%d%d", &u, &v); 104 G[u].push_back(v); 105 p[v] = u; 106 } 107 for (int i = 0; i < n; i++){ 108 if (!p[i]) { 109 dfs(i); 110 break; 111 } 112 } 113 build(1, n, 1); 114 for (int i = 0; i < n; i++){ 115 scanf("%d", &a); 116 cal(a,d[i]); 117 upd(L[i], 1, d[i]); 118 } 119 scanf("%d", &q); 120 while (q--){ 121 char s[10]; 122 int d2[6] = {0}; 123 scanf("%s%d", s, &a); 124 if (s[0] =='R'){ 125 memset(Ans, 0, sizeof Ans); 126 query(L[a], R[a], 1); 127 LL ans = 1; 128 LL num = 1; 129 for (int i = 0; i < 6; i++){ 130 //debug(Ans[i]); 131 num = (num*qpow(b[i], Ans[i]))%mod; 132 ans = ans*(Ans[i]+1)%mod; 133 } 134 printf("%lld %lld ", num, ans); 135 } 136 else { 137 int c; 138 scanf("%d", &c); 139 cal(c, d2); 140 upd(L[a], 1, d2); 141 } 142 } 143 return 0; 144 }
赛后微调（并没有什么改变）：202ms，27MB
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 #include<vector> 6 #include<queue> 7 #include<cmath> 8 #include<set> 9 #define lid id<<1 10 #define rid id<<1|1 11 #define INF 0x3f3f3f3f 12 #define LL long long 13 #define debug(x) cout << "[" << x << "]" << endl 14 using namespace std; 15 16 int b[] = {2, 3, 5, 7, 11, 13}; 17 const int mx = 1e5+10; 18 const int mod = 1e9+7; 19 int L[mx], R[mx], p[mx]; 20 struct tree{ 21 int l, r; 22 int p[6]; 23 int lazy[6]; 24 }tree[mx<<2]; 25 vector<int> G[mx]; 26 int cnt = 0; 27 LL Ans[6] = {0}; 28 29 void cal(int x, int *d){ 30 for(int i = 0; i < 6 ; i++){ 31 while(x % b[i] == 0){ 32 x /= b[i]; 33 d[i]++; 34 } 35 } 36 } 37 38 LL qpow(LL x, LL n){ //x^n 39 LL res = 1; 40 while (n > 0){ 41 if (n & 1) res = res*x%mod; 42 x = x*x % mod; 43 n >>= 1; 44 } 45 return res; 46 } 47 48 void push_up(int id){ 49 for (int i = 0; i < 6; i++) 50 tree[id].p[i] = tree[lid].p[i]+tree[rid].p[i]; 51 } 52 53 void build(int l, int r, int id){ 54 tree[id].l = l; 55 tree[id].r = r; 56 for (int i = 0; i < 6; i++) tree[id].p[i] = 0; 57 if (l == r) return; 58 int mid = (l+r) >> 1; 59 build(l, mid, lid); 60 build(mid+1, r, rid); 61 } 62 63 void dfs(int u){ 64 L[u] = ++cnt; 65 int len = G[u].size(); 66 for (int i = 0; i < len; i++){ 67 int v = G[u][i]; 68 dfs(v); 69 } 70 R[u] = cnt; 71 } 72 73 void upd(int c, int id, int *x){ 74 if (tree[id].l == c && tree[id].r == c){ 75 for (int i = 0; i < 6; i++) 76 tree[id].p[i] += x[i]; 77 return; 78 } 79 int mid = (tree[id].l + tree[id].r)>>1; 80 if (c <= mid) upd(c, lid, x); 81 else upd(c, rid, x); 82 push_up(id); 83 } 84 85 void query(int l, int r, int id){ 86 if (tree[id].l == l && tree[id].r == r){ 87 for (int i = 0; i < 6; i++) 88 Ans[i] += tree[id].p[i]; 89 return; 90 } 91 int mid = (tree[id].l + tree[id].r)>>1; 92 if (r <= mid) query(l, r, lid); 93 else if (mid < l) query(l, r, rid); 94 else { 95 query(l, mid, lid); 96 query(mid+1, r, rid); 97 } 98 } 99 100 int main(){ 101 int n, u, v, a, q; 102 scanf("%d", &n); 103 for (int i = 1; i < n; i++){ 104 scanf("%d%d", &u, &v); 105 G[u].push_back(v); 106 p[v] = u; 107 } 108 for (int i = 0; i < n; i++){ 109 if (!p[i]) { 110 dfs(i); 111 break; 112 } 113 } 114 build(1, n, 1); 115 for (int i = 0; i < n; i++){ 116 int d[6] = {0}; 117 scanf("%d", &a); 118 cal(a, d); 119 upd(L[i], 1, d); 120 } 121 scanf("%d", &q); 122 while (q--){ 123 char s[10]; 124 scanf("%s%d", s, &a); 125 if (s[0] == 'R'){ 126 memset(Ans, 0, sizeof Ans); 127 query(L[a], R[a], 1); 128 LL ans = 1, num = 1; 129 for (int i = 0; i < 6; i++){ 130 num = (num*qpow(b[i], Ans[i]))%mod; 131 ans = ans*(Ans[i]+1)%mod; 132 } 133 printf("%lld %lld ", num, ans); 134 } 135 else { 136 int c; 137 int d2[6] = {0}; 138 scanf("%d", &c); 139 cal(c, d2); 140 upd(L[a], 1, d2); 141 } 142 } 143 return 0; 144 }