HDU-5475 An easy problem(アナログ|(逆算+線分ツリー))
An easy problem
Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
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. (
0if 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
Sample Output
初期時x=1、毎回2種類の操作があります.
操作1:xに数を乗算
操作2:xをn回目の操作で割った数(nは1回のみ)
操作毎にx%modを出力する.
①最初から直接計算を考えていたが、タイムアウトすると思って諦めていたので、直接計算(乗算は直接乗せればよい;除法時に除法をマークしてから再計算すればよい)ができるとは思わなかった.
②后で大神の说明を闻いたところ、必ず乗る数があり、tmp[i]と计算し、后から除算した数n[i]を计算すると、ans[i]=(tmp[i]*n[i])%mod;約2300 ms;
③オオカミが自分で線分樹で作ったもので、1500 msくらいです.
Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
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
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
初期時x=1、毎回2種類の操作があります.
操作1:xに数を乗算
操作2:xをn回目の操作で割った数(nは1回のみ)
操作毎にx%modを出力する.
①最初から直接計算を考えていたが、タイムアウトすると思って諦めていたので、直接計算(乗算は直接乗せればよい;除法時に除法をマークしてから再計算すればよい)ができるとは思わなかった.
②后で大神の说明を闻いたところ、必ず乗る数があり、tmp[i]と计算し、后から除算した数n[i]を计算すると、ans[i]=(tmp[i]*n[i])%mod;約2300 ms;
#include <cstdio>
#define LL long long
using namespace std;
int n[100005],ope[100005],tmp[100005],ans[100005],x,mod;
bool flg[100005],mul[100005];//flg 1；mul
int main() {
int T,kase=0,Q,t,i,j;
scanf("%d",&T);
while(kase<T) {
printf("Case #%d:
",++kase);
scanf("%d%d",&Q,&mod);
tmp[0]=1;
for(i=1;i<=Q;++i) {
scanf("%d%d",&t,ope+i);
n[i]=1;
if(t==1)
mul[i]=flg[i]=true;
else
flg[i]=mul[i]=mul[ope[i]]=false;
}
for(i=1;i<=Q;++i) {
if(mul[i])
tmp[i]=((LL)tmp[i-1]*ope[i])%mod;
else
tmp[i]=tmp[i-1];
}
for(i=Q;i;--i) {
if(!flg[i]) {
x=ope[ope[i]];
for(j=ope[i];j<i;++j)//ope[i]~i-1 n[j] x
n[j]=((LL)n[j]*x)%mod;
}
ans[i]=((LL)n[i]*tmp[i])%mod;
}
for(i=1;i<=Q;++i)
printf("%d
",ans[i]);
}
return 0;
}
③オオカミが自分で線分樹で作ったもので、1500 msくらいです.