hdu 5475 An easy problem（暴力 || 线段树区间单点更新）

http://acm.hdu.edu.cn/showproblem.php?pid=5475

An easy problem

Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 755 Accepted Submission(s): 431

 

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≤10^5), 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≤10^5,1≤M≤10^9)

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≤10^9)
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

 

题目大意：t组测试数据，q个操作，每个操作（a, y）的结果X对m取余，X最初值为1，如果a==1，将X乘以y

如果a==2，将X除以第y次操作的数（即X要除的数曾出现在乘操作里）

 

线段树：线段树区间维护乘积，将q次操作当做q个节点，每个区间的初始值是1，乘操作（a，y）：将该次操作即该节点的数值改成其要乘的值y，

除操作（a，y）：将第y次操作即y节点的数值改成1（即可让其免去乘操作），输出从1到q区间的乘积即可

 

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
#define Lson root<<1, L, tree[root].Mid()
#define Rson root<<1|1, tree[root].Mid() + 1, R

const int N = 500010;
typedef long long ll;

struct Tree
{
    ll L, R;
    ll sum;
    int Mid()
    {
        return (L + R) / 2;
    }
} tree[N * 4];

ll a[N], m;

void Push(int root)
{
    tree[root].sum = (tree[root<<1].sum * tree[root<<1|1].sum) % m;
}//维护区间乘积

void Build(int root, ll L, ll R)
{
    tree[root].L = L, tree[root].R = R;
    if(L == R)
    {
        tree[root].sum = 1;
        return ;
    }

    Build(Lson);
    Build(Rson);

    Push(root);
}//建树

void Update(int root, ll op, ll e)
{
    if(tree[root].L == op && tree[root].R == op)
    {
        tree[root].sum = e % m;
        return ;
    }
    if(op <= tree[root].Mid())
        Update(root<<1, op, e);
    else
        Update(root<<1|1, op, e);
    Push(root);
}//区间单点更新

int main()
{
    int t, q, op, x = 0;
    scanf("%d", &t);
    while(t--)
    {
        x++;
        scanf("%d%lld", &q, &m);
        printf("Case #%d:\n", x);
        Build(1, 1, q);
        for(int i = 1 ; i <= q ; i++)
        {
            scanf("%d%lld", &op, &a[i]);
            if(op == 1)
                Update(1, i, a[i]);
            else
                Update(1, a[i], 1);
            printf("%lld\n", tree[1].sum);
        }
    }
    return 0;
}

 

 

通过用线段树的思路可以用暴力的方法来解题

 

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
#define Lson root<<1, L, tree[root].Mid()
#define Rson root<<1|1, tree[root].Mid() + 1, R

const int N = 500010;
typedef long long ll;

ll a[N], m;

int main()
{
    int t, op, q, x = 0;
    scanf("%d", &t);
    while(t--)
    {
        x++;
        ll ans = 1;
        scanf("%d%lld", &q, &m);
        printf("Case #%d:\n", x);
        for(int i = 1 ; i <= q ; i++)
        {
            scanf("%d%lld", &op, &a[i]);
            if(op == 1)
                ans = ans * a[i] % m;
            else
            {
                a[a[i]] = 1;
                a[i] = 1;
                ans = 1;
                for(int j = 1 ; j < i ; j++)
                    ans = ans * a[j] % m;
            }
            printf("%lld\n", ans);
        }
    }
    return 0;
}