FZU 2297 Number theory【線段樹/單點更新/思維】
Given a integers x = 1, you have to apply Q (Q ≤ 100000) operations: Multiply, Divide.
Input
First line of the input file contains an integer T(0 < T ≤ 10) that indicates how many cases of inputs are there.
The description of each case is given below:
The first line contains two integers Q and M. The next Q lines contains the operations in ith line following form:
M yi: x = x * yi.
N di: x = x / ydi.
It’s ensure that di is different. That means you can divide yi only once after yi came up.
0 < yi ≤ 10^9, M ≤ 10^9
Output
For each operation, print an integer (one per line) x % M.
Sample Input
1
10 1000000000
M 2
D 1
M 2
M 10
D 3
D 4
M 6
M 7
M 12
D 7
Sample Output
2
1
2
20
10
6
42
504
84
【分析】
針對一個數組的操作,即對一個區間。可以用線段樹去進行維護。初始化建樹,葉子節點的值為1,維護每段區間上各個元素的乘積sum。M yi,將第i個元素的值改為yi。N di,將第di個元素的值改為1。輸出即查詢區間[1,Q]的sum值。也就是變成了單點更新、區間查詢問題。
時間復雜度為O(QlongQ)。
#include<cstdio> #include<string> #include<cstdlib> #include<cmath> #include<iostream> #include<cstring> #include<set> #include<queue> #include<algorithm> #include<vector> #include<map> #include<stack> #include<sstream> #include<list> #include<bitset> using namespace std; typedef long long LL; typedef unsigned long long ULL; const int maxn = 1e5; const double eps = 1e-8; LL mod,val; LL sum[maxn*4]; void update(int p,int val,int l,int r,int rt) { if(l==r) { sum[rt]=val; return ; } int m=(l+r)/2; if(p<=m) update(p,val,l,m,rt*2); else update(p,val,m+1,r,rt*2+1); sum[rt] = sum[rt*2] * sum[rt*2+1] % mod; } //char op[10]; int main() { //ios::sync_with_stdio(false); int t,q; scanf("%d",&t); while(t--) { scanf("%d%lld",&q,&mod); for(int i=1;i<=4*maxn;i++) sum[i]=1; for(int i=1;i<=q;i++) { int x;char op[10]; scanf("%s%d",op,&x); if(op[0]=='M') { update(i,x,1,maxn,1); printf("%lld\n",sum[1]); } else { update(x,1,1,maxn,1); printf("%lld\n",sum[1]); } } } return 0; } /* 1 10 1000000000 M 2 D 1 M 2 M 10 D 3 D 4 M 6 M 7 M 12 D 7 2 1 2 20 10 1 6 42 504 84 */
FZU 2297 Number theory【線段樹/單點更新/思維】