HDU 5475An easy problem 離線set/線段樹

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An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem DescriptionOne 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. 

 

InputThe 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.
 

 

OutputFor 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 Input110 10000000001 22 11 21 102 32 41 61 71 122 7 

 

Sample OutputCase #1:2122010164250484 

 

Source2015 ACM/ICPC Asia Regional Shanghai Online  題解：考慮到取摸，離線做或者線段樹上搞

///1085422276#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<queue>#include<cmath>#include<map>#include<bitset>#include<set>#include<vector>using namespace std ;typedef long long ll;#define mem(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,127,sizeof(a));#define TS printf("111111\n");#define FOR(i,a,b) for( int i=a;i<=b;i++)#define FORJ(i,a,b) for(int i=a;i>=b;i--)#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)#define mod 1000000007#define inf 100000inline ll read(){    ll x=0,f=1;    char ch=getchar();    while(ch<‘0‘||ch>‘9‘)    {        if(ch==‘-‘)f=-1;        ch=getchar();    }    while(ch>=‘0‘&&ch<=‘9‘)    {        x=x*10+ch-‘0‘;        ch=getchar();    }    return x*f;}//****************************************struct ss{    int id,x;    bool operator < (const ss &A)const    {        return id < A.id;    }};#define maxn 100000+5set<ss >s;set<ss >::iterator it;int main(){    int oo=1;    ll n,q,m,x[maxn],op[maxn],vis[maxn],ans[maxn],A[maxn];    int T=read();    while(T--)    {        scanf("%I64d%I64d",&n,&m);        FOR(i,1,n)        {            scanf("%I64d%I64d",&op[i],&x[i]);        }        mem(vis);        FOR(i,1,n)        {            if(op[i]==2)vis[x[i]]=1;        }        mem(ans);        ans[0]=1;        FOR(i,1,n)        {            ans[i]=ans[i-1];            if(op[i]==1&&!vis[i])            {                ans[i]=(ans[i]*x[i])%m;            }        }        //cout<<ans[10]<<endl;        s.clear();        for(int i=n; i>=1; i--)        {            ll tmp=1;            for(it=s.begin(); it!=s.end(); it++)            {                if((*it).id>i)break;                tmp*=(*it).x;                tmp%=m;            }            A[i]=(ans[i]*tmp)%m;            if(op[i]==2)            {                ss g;                g.id=x[i];                g.x=x[x[i]];                s.insert(g);            }        }        printf("Case #%d:\n",oo++);        for(int i=1; i<=n; i++)            cout<<A[i]<<endl;    }    return 0;}

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HDU 5475An easy problem 離線set/線段樹