杭电5475/2015 ACM/ICPC Asia Regional Shanghai Online An easy problem（线段树过）

An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1137    Accepted Submission(s): 557

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

110 10000000001 22 11 21 102 32 41 61 71 122 7

 

Sample Output

Case #1:2122010164250484

这是一道怎么说的题目呢。刚开始看到5000MS，直接暴力去做。

然后我得到了这样的一个思路：（暴力思路）

一步一步乘啊乘，乘啊乘，除啊除，除啊除。

然后感觉好二，因为我是和我队友一起做，他跟我提出个道理，除掉哪组数据，就相当于不乘这组数据。因为毕竟除数据是会有纰漏的，会有精度消耗之类的不可避免的问题。然后还是暴力解：乘啊乘乘啊乘，标记啊标记，标记啊标记。然后就TLE了~

后来发现，如果是求数据乘积，我可以利用线段树去做啊~然后就开始构思：因为是n组数据，一共所有数也就n个，因为是乘积，所以我初始化为1就行：

void pushup(int rt){    tree[rt]=(tree[rt<<1]%m*tree[rt<<1|1]%m)%m;//别忘记求余}void build( int l ,int r , int rt ){if( l == r ){tree[rt]=1;return ;}else{int m = (l+r)>>1 ;build(lson) ;build(rson) ;pushup(rt) ;}}

然后是数据修改:如果是操作1，表示要乘上这个数，对应树上的位子改变数据为要乘的数就行，相反的如果是操作2，表示有某一组数据相当于不乘，那么相对应的位子改为乘1就行了：

void update(int p,int c,int l,int r,int rt)//p表示位子，c表示更改数据{    if(l==r)    {        tree[rt]=c;    }    else    {        int m=(l+r)>>1;        if(p<=m) update(p,c,lson);        else  update(p,c,rson);        pushup(rt);    }}

因为是求树中所有数的乘积，那么我们输出tree【1】就行了，这里是完整AC代码：

#include<stdio.h>#include<string.h>using namespace std;#define lson l,m,rt*2#define rson m+1,r,rt*2+1__int64 tree[1212121];int n,m;void pushup(int rt){    tree[rt]=(tree[rt<<1]%m*tree[rt<<1|1]%m)%m;}void build( int l ,int r , int rt ){if( l == r ){tree[rt]=1;return ;}else{int m = (l+r)>>1 ;build(lson) ;build(rson) ;pushup(rt) ;}}void update(int p,int c,int l,int r,int rt){    if(l==r)    {        tree[rt]=c;    }    else    {        int m=(l+r)>>1;        if(p<=m) update(p,c,lson);        else  update(p,c,rson);        pushup(rt);//printf("sum[%d]:%d\n",rt,tree[rt]);    }}int main(){    int t;    int kase=0;    scanf("%d",&t);    while(t--)    {        memset(tree,0,sizeof(tree));        scanf("%d%d",&n,&m);        build(1,n,1);        printf("Case #%d:\n",++kase);        for(int i=0;i<n;i++)        {            int q,c;            scanf("%d%d",&q,&c);            if(q==1)            {                update(i+1,c,1,n,1);                printf("%I64d\n",tree[1]%m);            }            if(q==2)            {                update(c,1,1,n,1);                printf("%I64d\n",tree[1]%m);            }        }    }}