hdu 5475 An easy problem 线段树
2015-10-03 18:08
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An easy problem
Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 746 Accepted Submission(s): 427
Problem Description
One day, a useless calculator was being built by Kuros. Let's assume that number X is showed on the screen of calculator. At first, X = 1. This calculator only supports two types of operation.
1. multiply X with a number.
2. divide X with a number which was multiplied before.
After each operation, please output the number X modulo M.
Input
The first line is an integer T(1≤T≤10),
indicating the number of test cases.
For each test case, the first line are two integers Q and M. Q is the number of operations and M is described above. (1≤Q≤105,1≤M≤109)
The next Q lines, each line starts with an integer x indicating the type of operation.
if x is 1, an integer y is given, indicating the number to multiply. (0<y≤109)
if x is 2, an integer n is given. The calculator will divide the number which is multiplied in the nth operation. (the nth operation must be a type 1 operation.)
It's guaranteed that in type 2 operation, there won't be two same n.
Output
For each test case, the first line, please output "Case #x:" and x is the id of the test cases starting from 1.
Then Q lines follow, each line please output an answer showed by the calculator.
Sample Input
1
10 1000000000
1 2
2 1
1 2
1 10
2 3
2 4
1 6
1 7
1 12
2 7
Sample Output
Case #1:
2
1
2
20
10
1
6
42
504
84
Source
2015 ACM/ICPC Asia Regional Shanghai Online
用线段树维护区间乘积即可。删除操作就是把数字置为1.
#include<iostream> #include<cstring> #include<algorithm> #include<cstdio> #include<vector> #include<map> using namespace std; #define maxn 300007 int M; int num[maxn],lc[maxn],rc[maxn],cnt; void build(int u,int l,int r){ num[u] = 1; if(l == r) return ; int mid = (l+r)/2; lc[u] = ++cnt; rc[u] = ++cnt; build(lc[u],l,mid); build(rc[u],mid+1,r); } void add(int u,int l,int r,int p,int val){ if(l == r){ num[u] = val; return ; } int mid = (l+r)/2; if(p > mid) add(rc[u],mid+1,r,p,val); else add(lc[u],l,mid,p,val); num[u] = ((long long)num[lc[u]]*num[rc[u]])%M; } int main(){ int t,n,tt=1,u,v; scanf("%d",&t); while(t--){ scanf("%d%d",&n,&M); cnt = 0; build(0,0,n-1); printf("Case #%d:\n",tt++); for(int i = 0;i < n; i++){ scanf("%d%d",&u,&v); if(u == 1) add(0,0,n-1,i,v); else add(0,0,n-1,v-1,1); printf("%d\n",num[0]%M); } } return 0; }
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