An easy problem
Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 467 Accepted Submission(s): 258
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.
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.
For each test case, the first line are two integers Q and M. Q is the number of operations and M is described above. (
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. (
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.
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
题意是一台计算器,最开始的数字是1,然后对其不断操作,输入的操作为1时,乘以后面的数y。输入的操作为2时,除以第y次操作的数。问每一次操作后的结果是多少。
哇,这题居然是线段树做法,真的是没想到啊,发现线段树的题竟然这么广。
对每一个定点进行更新,对整个线段树求乘积。
代码:
#include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <string> #include <cstring> #pragma warning(disable:4996) using namespace std; struct no { int L,R; long long mul; }tree[400015]; int root,n; long long mod; void buildtree(int root,int L,int R) { tree[root].L=L; tree[root].R=R; tree[root].mul=1; if(L!=R) { buildtree(root*2+1,L,(L+R)/2); buildtree(root*2+2,(L+R)/2+1,R); } } void insert(int root,int s,int e,long long val) { if(tree[root].L==s&&tree[root].R==e) { tree[root].mul=val%mod; return; } if(e<=(tree[root].L+tree[root].R)/2) { insert(root*2+1,s,e,val); } else if(s>=(tree[root].L+tree[root].R)/2+1) { insert(root*2+2,s,e,val); } else { insert(root*2+1,s,(tree[root].L+tree[root].R)/2,val); insert(root*2+2,(tree[root].L+tree[root].R)/2+1,e,val); } tree[root].mul = (tree[2*root+1].mul * tree[2*root+2].mul)%mod; } int main() { //freopen("i.txt","r",stdin); //freopen("o.txt","w",stdout); int test,i,j,oper; long long val; scanf("%d",&test); for(i=1;i<=test;i++) { printf("Case #%d: ",i); scanf("%d%lld",&n,&mod); buildtree(0,1,n); for(j=1;j<=n;j++) { scanf("%d%lld",&oper,&val); if(oper==1) { insert(0,j,j,val); } else { insert(0,val,val,1); } printf("%lld ",tree[0].mul); } } //system("pause"); return 0; }
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